Geometrical Optics
272 Questions
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th July Evening Shift
In the given figure, the face $A C$ of the equilateral prism is immersed in a liquid of refractive index '$n$'. For incident angle $60^{\circ}$ at the side $A C$, the refractive light beam just grazes along face $A C$. The refractive index of the liquid $n=\frac{\sqrt{x}}{4}$. The value of $x$ is ____________.
(Given refractive index of glass $=1.5$ )
Correct Answer: 27
Explanation:
Given prism is equilateral so, angel of prism, A $=60^{\circ}$
On first surface light is passing straight without any deviation, hence angle of refraction $\mathrm{r}_1=0^{\circ}$.
$ \begin{aligned} & r_1+i_2=\mathrm{A} \\\\ & 0+i_2=60^{\circ} \\\\ & \Rightarrow i_2=60^{\circ} \end{aligned} $
On $2^{\text {nd }}$ surface ray, refracting ray is becoming parallel to the junction of surfaces, so $i_2$ will act as critical angle.
From snell's law $\mu_1 \sin \theta_1=\mu_2 \sin \theta_2$
$ \begin{aligned} & \Rightarrow \frac{3}{2} \sin i_2=\mu \sin \theta_2 \\\\ & \Rightarrow \mu=\frac{3}{2} \times \frac{\sin 60^{\circ}}{\sin 90^{\circ}}=\frac{3}{2} \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4} \text { or } \frac{\sqrt{27}}{4} \end{aligned} $
$\Rightarrow$ Comparing above equation with
given $n=\frac{\sqrt{x}}{4}$, we get $x=27$
On first surface light is passing straight without any deviation, hence angle of refraction $\mathrm{r}_1=0^{\circ}$.
$ \begin{aligned} & r_1+i_2=\mathrm{A} \\\\ & 0+i_2=60^{\circ} \\\\ & \Rightarrow i_2=60^{\circ} \end{aligned} $
On $2^{\text {nd }}$ surface ray, refracting ray is becoming parallel to the junction of surfaces, so $i_2$ will act as critical angle.
From snell's law $\mu_1 \sin \theta_1=\mu_2 \sin \theta_2$
$ \begin{aligned} & \Rightarrow \frac{3}{2} \sin i_2=\mu \sin \theta_2 \\\\ & \Rightarrow \mu=\frac{3}{2} \times \frac{\sin 60^{\circ}}{\sin 90^{\circ}}=\frac{3}{2} \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4} \text { or } \frac{\sqrt{27}}{4} \end{aligned} $
$\Rightarrow$ Comparing above equation with
given $n=\frac{\sqrt{x}}{4}$, we get $x=27$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th July Morning Shift
The graph between $\frac{1}{u}$ and $\frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is $1.5$ and its both the surfaces have same radius of curvature $R$. The value of $R$ will be ____________ $\mathrm{cm} .$ (where $u=$ object distance, $v=$ image distance)
Correct Answer: 10
Explanation:
$f = 10$ cm
${1 \over f} = (\mu - 1)\left( {{1 \over R} - {1 \over { - R}}} \right)$
${1 \over {10}} = {{1.5 - 1} \over 1} \times {2 \over R}$
$ {1 \over {10}} = {1 \over R}$
$R = 10$ cm
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th July Evening Shift
A convex lens of focal length 20 cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60 cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is ____________ cm.
Correct Answer: 10
Explanation:
${1 \over v} + {1 \over u} = {1 \over f}$
${1 \over v} - {1 \over { - 60}} = {1 \over {20}}$
${1 \over v} = - {1 \over {60}} + {1 \over {20}} = {{ - 1 + 3} \over {60}} = {2 \over {60}}$
$ \Rightarrow v = \, +\, 30$ cm
$\therefore$ Radius of curvature of mirror = 30 $-$ 10 = 20 cm
$ \Rightarrow {f_{mirror}} = {{20} \over 2} = 10$ cm
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 30th June Morning Shift
The refractive index of an equilateral prism is $\sqrt 2 $. The angle of emergence under minimum deviation position of prism, in degree, is ___________.
Correct Answer: 45
Explanation:
Refractive index
$ \begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta \sin }{2}\right)}{\sin \left(\frac{A}{2}\right)} \Rightarrow \sqrt{2}=\sin \frac{\left(\frac{60+\delta \min }{2}\right)}{\sin 30^{\circ}} \\\\ & \Rightarrow \frac{1}{2}=\sin \left(\frac{60+\delta \min }{2}\right) \Rightarrow 45^{\circ}=\frac{60+\delta \min }{2} \\\\ & \Rightarrow \delta \min =30^{\circ} \\\\ & \delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \\\\ & \text { Here, } e=i \\\\ & \text { So, } \delta \mathrm{min}=2 \mathrm{e}-\mathrm{A} \Rightarrow 2 \mathrm{e}=\delta \min +\mathrm{A} \\\\ & e=\frac{\delta \min +A}{2}=\frac{30^{\circ}+60^{\circ}}{2}=45^{\circ} \end{aligned} $
$ \begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta \sin }{2}\right)}{\sin \left(\frac{A}{2}\right)} \Rightarrow \sqrt{2}=\sin \frac{\left(\frac{60+\delta \min }{2}\right)}{\sin 30^{\circ}} \\\\ & \Rightarrow \frac{1}{2}=\sin \left(\frac{60+\delta \min }{2}\right) \Rightarrow 45^{\circ}=\frac{60+\delta \min }{2} \\\\ & \Rightarrow \delta \min =30^{\circ} \\\\ & \delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \\\\ & \text { Here, } e=i \\\\ & \text { So, } \delta \mathrm{min}=2 \mathrm{e}-\mathrm{A} \Rightarrow 2 \mathrm{e}=\delta \min +\mathrm{A} \\\\ & e=\frac{\delta \min +A}{2}=\frac{30^{\circ}+60^{\circ}}{2}=45^{\circ} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 29th June Morning Shift
A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is _____________ mm.
Correct Answer: 225
Explanation:
1st refraction : ${{1.5} \over {{v_1}}} - 0 = {{0.5} \over {15}}$
$\Rightarrow$ v1 = 45 cm
2nd refraction : ${1 \over {{v_2}}} - {{1.5} \over {15}} = {{ - 0.5} \over { - 15}}$
$ \Rightarrow {1 \over {{v_2}}} = {1 \over {30}} + {1 \over {10}}$
$ = {4 \over {30}}$
$\Rightarrow$ v2 = + 7.5 cm
$\Rightarrow$ Distance from centre = 22.5 cm = 225 mm
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Evening Shift
A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt7$ m. The refractive index of water is ${4 \over 3}$. The area of the surface of water through which light from the bulb can emerge out is x$\pi$ m2. The value of x is __________.
Correct Answer: 9
Explanation:
So $r = h{{\sin {i_c}} \over {\sqrt {1 - {{\sin }^2}{i_c}} }}$
So $A = \pi {r^2}$
$ = {{\pi {h^2}{{\sin }^2}{i_c}} \over {1 - {{\sin }^2}{i_c}}}$
$ = {{\pi 7 \times {9 \over {16}}} \over {1 - {9 \over {16}}}} = {{\pi \times 7 \times 9} \over 7} = 9\pi $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Morning Shift
A light ray is incident, at an incident angle $\theta$1, on the system of tow plane mirrors M1 and M2 having an inclination angle 75$^\circ$ between them (as shown in figure). After reflecting from mirror M1 it gets reflected back by the mirror M2 with an angle of reflection 30$^\circ$. The total deviation of the ray will be _____________ degree.
Correct Answer: 210
Explanation:
On first reflection angel of deviation is 90° and on second reflection angle of deviation is 120°
So total deviation is $\delta $ = 90° +120° = 210°
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Evening Shift
A ray of light is incident at an angle of incidence 60$^\circ$ on the glass slab of refractive index $\sqrt3$. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is 4$\sqrt3$ cm. The thickness of the glass slab is __________ cm.
Correct Answer: 12
Explanation:
$1 \times \sin 60^\circ = \sqrt 3 \times \sin r$
$ \Rightarrow r = 30^\circ $
$\therefore$ ${l_1} = 4\sqrt 3 \times 2$
$ = 8\sqrt 3 $ cm
$\therefore$ Thickness, $t = {l_1}\cos 30^\circ $
$ = 8\sqrt 3 \times {{\sqrt 3 } \over 2}$
$ = 4 \times 3$
$ = 12$ cm
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Morning Shift
Two identical thin biconvex lens of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is __________ cm.
Correct Answer: 10
Explanation:
${1 \over {{f_l}}} = \left( {{{{\mu _e}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$
here $|{R_1}| = |{R_2}| = R$
$ \Rightarrow {1 \over {{f_{{l_1}}}}} = (1.5 - 1)\left( {{2 \over R}} \right) = {1 \over {15}}$
$ \Rightarrow {1 \over R} = {1 \over {15}}$ or $R = 15$ cm
for the concave lens made up of liquid
${1 \over {{f_{{l_2}}}}} = (1.25 - 1)\left( { - {2 \over R}} \right) = - {1 \over {30}}$ cm
now for equivalent lens
${1 \over {{f_e}}} = {2 \over {{f_{{l_1}}}}} + {1 \over {{f_{{l_2}}}}}$
$ = {2 \over {15}} - {1 \over {30}} = {3 \over {30}} = {1 \over {10}}$
or ${f_e} = 10$ cm
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water ($\mu$ = 4/3) into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?
A.
11.7 cm
B.
10 cm
C.
7.5 cm
D.
8.75 cm
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
An object is placed at the focus of concave lens having focal length f. What is the magnification and distance of the image from the optical centre of the lens?
A.
1, $\infty$
B.
Very high, $\infty$
C.
${1 \over 2}$, ${f \over 2}$
D.
${1 \over 4}$, ${f \over 4}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Morning Shift
Two plane mirrors M1 and M2 are at right angle to each other shown. A point source 'P' is placed at 'a' and '2a' meter away from M1 and M2 respectively. The shortest distance between the images thus formed is : (Take $\sqrt 5 $ = 2.3)
A.
3a
B.
4.6a
C.
2.3a
D.
2$\sqrt {10} $a
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Evening Shift
Curved surfaces of a plano-convex lens of refractive index $\mu$1 and a plano-concave lens of refractive index $\mu$2 have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.
A.
${1 \over {{\mu _2} - {\mu _1}}}$
B.
${\mu _1} - {\mu _2}$
C.
${1 \over {{\mu _1} - {\mu _2}}}$
D.
${\mu _2} - {\mu _1}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Morning Shift
An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is :
A.
${{2{d_1}{d_2}} \over {{d_1} - {d_2}}}$
B.
${{2{d_1}{d_2}} \over {{d_1} + {d_2}}}$
C.
${{{d_1}{d_2}} \over {{d_1} + {d_2}}}$
D.
${{{d_1}{d_2}} \over {{d_1} - {d_2}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th August Morning Shift
Find the distance of the image from object O, formed by the combination of lenses in the figure :
A.
75 cm
B.
10 cm
C.
20 cm
D.
infinity
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
Car B overtakes another car A at a relative speed of 40 ms$-$1. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?
A.
4 ms$-$1
B.
0.2 ms$-$1
C.
40 ms$-$1
D.
0.1 ms$-$1
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Evening Shift
The expected graphical representation of the variation of angle of deviation '$\delta$' with angle of incidence 'i' in a prism is :
A.
B.
C.
D.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
A prism of refractive index $\mu$ and angle of prism A is placed in the position of minimum angle of deviation. If minimum angle of deviation is also A, then in terms of refractive index
A.
$2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$
B.
${\sin ^{ - 1}}\left( {{\mu \over 2}} \right)$
C.
${\sin ^{ - 1}}\left( {\sqrt {{{\mu - 1} \over 2}} } \right)$
D.
${\cos ^{ - 1}}\left( {{\mu \over 2}} \right)$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
A ray of light entering from air into a denser medium of refractive index ${4 \over 3}$, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle $\theta$ should be equal to :
A.
${\sin ^{ - 1}}{{\sqrt 7 } \over 3}$
B.
${\sin ^{ - 1}}{{\sqrt 5 } \over 4}$
C.
${\sin ^{ - 1}}{{\sqrt 7 } \over 4}$
D.
${\sin ^{ - 1}}{{\sqrt 5 } \over 3}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Morning Shift
A ray of laser of a wavelength 630 nm is incident at an angle of 30$^\circ$ at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is 2.42 and that of air is 1. Choose the correct option.
A.
angle of refraction is 24.41$^\circ$
B.
angle of refraction is 30$^\circ$
C.
refraction is not possible
D.
angle of refraction is 53.4$^\circ$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
A ray of light passes from a denser medium to a rarer medium at an angle of incidence i. The reflected and refracted rays make an angle of 90$^\circ$ with each other. The angle of reflection and refraction are respectively r and r'. The critical angle is given by
A.
sin$-$1 (tan r)
B.
sin$-$1 (cot r)
C.
sin$-$1 (tan r')
D.
tan$-$1 (sin i)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Morning Shift
Region I and II are separated by a spherical surface of radius 25 cm. An object is kept in region I at a distance of 40 cm from the surface. The distance of the image from the surface is :
A.
55.44 cm
B.
18.23 cm
C.
9.52 cm
D.
37.58 cm
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Evening Shift
Three rays of light, namely red (R), green (G) and blue (B) are incident on the face PQ of a right angled prism PQR as shown in the figure.
The refractive indices of the material of the prism for red, green and blue wavelength are 1.27, 1.42 and 1.49 respectively. The colour of the ray(s) emerging out of the face PR is :
The refractive indices of the material of the prism for red, green and blue wavelength are 1.27, 1.42 and 1.49 respectively. The colour of the ray(s) emerging out of the face PR is :
A.
green
B.
blue and green
C.
blue
D.
red
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 18th March Morning Shift
Your friend is having eye sight problem. She is not able to see clearly a distant uniform window mesh and it appears to her as non-uniform and distorted. The doctor diagnosed the problem as :
A.
Astigmatism
B.
Myopia with Astigmatism
C.
Presbyopia with Astigmatism
D.
Myopia and Hypermetropia
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
The thickness at the centre of a plane convex lens is 3 mm and the diameter is 6 cm. If the speed of light in the material of the lens is 2 $\times$ 108 ms$-$1. The focal length of the lens is ____________.
A.
0.30 cm
B.
30 cm
C.
15 cm
D.
1.5 cm
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R1 and R2 respectively.
A.
Zero
B.
Infinite
C.
1
D.
${{{R_1}{R_2}} \over {{R_1} - {R_2}}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
Red light differs from blue light as they have :
A.
Different frequencies and same wavelengths
B.
Different frequencies and different wavelengths
C.
Same frequencies and different wavelengths
D.
Same frequencies and same wavelengths
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
The angle of deviation through a prism is minimum when
(A) Incident ray and emergent ray are symmetric to the prism.
(B) The refracted ray inside the prism becomes parallel to its base
(C) Angle of incidence is equal to that of the angle of emergence
(D) When angle of emergence is double the angle of incidence
Choose the correct answer from the options given below :
(A) Incident ray and emergent ray are symmetric to the prism.
(B) The refracted ray inside the prism becomes parallel to its base
(C) Angle of incidence is equal to that of the angle of emergence
(D) When angle of emergence is double the angle of incidence
Choose the correct answer from the options given below :
A.
Statements (B) and (C) are true
B.
Only statements (A) and (B) are true
C.
Statements (A), (B) and (C) are true
D.
Only statement (D) is true
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ respectively. Then choose the correct relation for these vectors.
A.
$\overrightarrow b $ = $\overrightarrow a $ + 2$\overrightarrow c $
B.
$\overrightarrow b $ = $\overrightarrow a $ $-$ 2 ($\overrightarrow a $ . $\overrightarrow c $)$\overrightarrow c $
C.
$\overrightarrow b $ = 2$\overrightarrow a $ + $\overrightarrow c $
D.
$\overrightarrow b $ = $\overrightarrow a $ $-$ $\overrightarrow c $
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
Given below are two statements : one is labeled as Assertion A and the other is labeled as Reason R.
Assertion A : For a simple microscope, the angular size of the object equals the angular size of the image.
Reason R : Magnification is achieved as the small object can be kept much closer to the eye than 25 cm and hence it subtends a large angle.
In the light of the above statements, choose the most appropriate answer from the options given below :
Assertion A : For a simple microscope, the angular size of the object equals the angular size of the image.
Reason R : Magnification is achieved as the small object can be kept much closer to the eye than 25 cm and hence it subtends a large angle.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
A is true but R is false
B.
A is false but R is true
C.
Both A and R are true and R is the correct explanation of A
D.
Both A and R are true but R is NOT the correct explanation of A
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
A short straight object of height 100 cm lies before the central axis of a spherical mirror whose focal length has absolute value | f | = 40 cm. The image of object produced by the mirror is of height 25 cm and has the same orientation of the object. One may conclude from the information :
A.
Image is real, same side of convex mirror.
B.
Image is virtual, opposite side of convex mirror.
C.
Image is virtual, opposite side of concave mirror.
D.
Image is real, same side of concave mirror.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
The focal length f is related to the radius of curvature r of the spherical convex mirror by :
A.
f = r
B.
f = $-$ r
C.
f = +${{1 \over 2}}$ r
D.
f = $-$${{1 \over 2}}$ r
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
Cross-section view of a prism is the equilateral triangle ABC in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from P (midpoint of BC) to A is ______________ $\times$ 10$-$10 s. (Given, speed of light in vacuum = 3 $\times$ 108 m/s and cos30$^\circ$ = ${{\sqrt 3 } \over 2}$)
Correct Answer: 5
Explanation:
i = A = 60$^\circ$
${{ \delta } _{\min }}$ = 2i $-$ A
= 2 $\times$ 60$^\circ$ $-$ 60$^\circ$ = 60$^\circ$
$\mu = {{{{\sin }^{ - 1}}\left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {{{\sin }^{ - 1}}\left( {{A \over 2}} \right)}}$
$ = \sqrt 3 $
Vprism $ = {{3 \times {{10}^8}} \over {\sqrt 3 }}$
AP = 10 $\times$ 10$-$2 $\times$ ${{\sqrt 3 } \over 2}$
time $ = {{5 \times {{10}^{ - 2}}} \over {3 \times {{10}^8}}} \times \sqrt 3 \times \sqrt 3 $
= 5 $\times$ 10$-$10 sec
Ans. = 5
${{ \delta } _{\min }}$ = 2i $-$ A
= 2 $\times$ 60$^\circ$ $-$ 60$^\circ$ = 60$^\circ$
$\mu = {{{{\sin }^{ - 1}}\left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {{{\sin }^{ - 1}}\left( {{A \over 2}} \right)}}$
$ = \sqrt 3 $
Vprism $ = {{3 \times {{10}^8}} \over {\sqrt 3 }}$
AP = 10 $\times$ 10$-$2 $\times$ ${{\sqrt 3 } \over 2}$
time $ = {{5 \times {{10}^{ - 2}}} \over {3 \times {{10}^8}}} \times \sqrt 3 \times \sqrt 3 $
= 5 $\times$ 10$-$10 sec
Ans. = 5
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
An object is placed at a distance of 12 cm from a convex lens. A convex mirror of focal length 15 cm is placed on other side of lens at 8 cm as shown in the figure. Image of object coincides with the object.
When the convex mirror is removed, a real and inverted image is formed at a position. The distance of the image from the object will be .............. (cm)
When the convex mirror is removed, a real and inverted image is formed at a position. The distance of the image from the object will be .............. (cm)
Correct Answer: 50
Explanation:
For the object to coincide with image, the light must fall perpendicularly to mirror. Which means that the light will have to converge at C of mirror. Without the mirror also, the light would coverage at C.
So the distance is : 12 + 8 + 30 = 50 cm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
A prism of refractive index n1 and another prism of refractive index n2 are stuck together (as shown in the figure). n1 and n2 depend on $\lambda$, the wavelength of light, according to the relation
${n_1} = 1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$ and ${n_2} = 1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$
The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be _____________ nm.
${n_1} = 1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$ and ${n_2} = 1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$
The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be _____________ nm.
Correct Answer: 600
Explanation:
For no bending, n1 = n2
$1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$ = $1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$
On solving,
9 $\times$ 10$-$14 = 25$\lambda$2
$\lambda$ = 6 $\times$ 10$-$7
$\lambda$ = 600 nm
$1.2 + {{10.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$ = $1.45 + {{1.8 \times {{10}^{ - 14}}} \over {{\lambda ^2}}}$
On solving,
9 $\times$ 10$-$14 = 25$\lambda$2
$\lambda$ = 6 $\times$ 10$-$7
$\lambda$ = 600 nm
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
A ray of light passing through a prism ($\mu$ = $\sqrt 3 $) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is _____________ (in degrees).
Correct Answer: 60
Explanation:
For minimum deviation r1 = r2 = A/2
given i = 2r
$\mu = {{\sin i} \over {\sin r}} = {{\sin 2r} \over {\sin r}}$
$ \Rightarrow \cos r = {\mu \over 2}$
$\Rightarrow$ r = 30$^\circ$
$\Rightarrow$ A = 60$^\circ$
given i = 2r
$\mu = {{\sin i} \over {\sin r}} = {{\sin 2r} \over {\sin r}}$
$ \Rightarrow \cos r = {\mu \over 2}$
$\Rightarrow$ r = 30$^\circ$
$\Rightarrow$ A = 60$^\circ$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to __________ cm.
Correct Answer: 25
Explanation:
Given, magnification, M = 6
Since we know that magnifying power of a simple microscope is given by
$M = 1 + {D \over {{f_0}}}$
where, D = least distance of distinct vision = 25 cm
and f0 = focal length of objective lens.
$ \Rightarrow 6 = 1 + {D \over {{f_0}}} \Rightarrow 6 = 1 + {{25} \over {{f_0}}} \Rightarrow 5 = {{25} \over {{f_0}}} \Rightarrow {f_0} = 5$ cm
For compound microscope, magnifying power is given by
$M = {{I\,.\,D} \over {{f_0}{f_e}}} = 2$Msimple microscope
where, f0 = fe are the focal lengths of the objective lens and eye piece respectively
and l = length of the given tube = 0.6 m
$ \Rightarrow 12 = {{60 \times 25} \over {5\,.\,{f_e}}}$ [$\because$ magnification is doubled]
$\Rightarrow$ fe = 25 cm
This is the required focal length of eyepiece.
Since we know that magnifying power of a simple microscope is given by
$M = 1 + {D \over {{f_0}}}$
where, D = least distance of distinct vision = 25 cm
and f0 = focal length of objective lens.
$ \Rightarrow 6 = 1 + {D \over {{f_0}}} \Rightarrow 6 = 1 + {{25} \over {{f_0}}} \Rightarrow 5 = {{25} \over {{f_0}}} \Rightarrow {f_0} = 5$ cm
For compound microscope, magnifying power is given by
$M = {{I\,.\,D} \over {{f_0}{f_e}}} = 2$Msimple microscope
where, f0 = fe are the focal lengths of the objective lens and eye piece respectively
and l = length of the given tube = 0.6 m
$ \Rightarrow 12 = {{60 \times 25} \over {5\,.\,{f_e}}}$ [$\because$ magnification is doubled]
$\Rightarrow$ fe = 25 cm
This is the required focal length of eyepiece.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at ${{{2^{rd}}} \over 3}$ of the distance of the object from the surface. The wavelength of light inside the surface is ${2 \over 3}$ times the wavelength in air. The radius of the curved surface is ${x \over {13}}$ m. The value of 'x' is ___________.
Correct Answer: 30
Explanation:
${{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{\mu _2} - {\mu _1}} \over R}$
${\mu _1} = 1$, ${\mu _2} = 1.5$
${{1.5} \over { + 10}} - {1 \over { - 15}} = {{1.5 - 1} \over { + R}}$
$R = {{30} \over {13}}$ m
${\mu _1} = 1$, ${\mu _2} = 1.5$
${{1.5} \over { + 10}} - {1 \over { - 15}} = {{1.5 - 1} \over { + R}}$
$R = {{30} \over {13}}$ m
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
A deviation of 2$^\circ$ is produced in the yellow ray when prism of crown and flint glass are achromatically combined. Taking dispersive powers of crown and flint glass as 0.02 and 0.03 respectively and refractive index for yellow light for these glasses are 1.5 and 1.6 respectively. The refracting angles for crown glass prism will be ____________$^\circ$ (in degree). (Round off to the Nearest Integer)
Correct Answer: 12
Explanation:
${\delta _{net}} = ({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2}$
$2^\circ = ({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2}$ ....... (1)
and ${\omega _1}({\mu _1} - 1){A_1} = {\omega _2}({\mu _2} - 1){A_2}$ ..... (2)
Substituting the values in equation (1) and (2), we get
$2^\circ = 0.5{A_1} - 0.6{A_2}$ ...... (3)
$10{A_1} = 18{A_2}$ ...... (4)
From equation (3) and (4)
$2^\circ = {{{A_1}} \over 2} - {{{A_1}} \over 3}$
${A_1} = 12^\circ $
$2^\circ = ({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2}$ ....... (1)
and ${\omega _1}({\mu _1} - 1){A_1} = {\omega _2}({\mu _2} - 1){A_2}$ ..... (2)
Substituting the values in equation (1) and (2), we get
$2^\circ = 0.5{A_1} - 0.6{A_2}$ ...... (3)
$10{A_1} = 18{A_2}$ ...... (4)
From equation (3) and (4)
$2^\circ = {{{A_1}} \over 2} - {{{A_1}} \over 3}$
${A_1} = 12^\circ $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
A point source of light S, placed at a distance 60cm in front of the centre of a plane mirror of width 50 cm, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 1.2 m from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is __________ cm.
Correct Answer: 150
Explanation:
Given, length of mirror, m = 50 cm = 50 $\times$ 10$-$2 m
Distance of source from mirror, d = 60 cm = 60 $\times$ 10$-$2 m
Distance of man from mirror, dm = 1.2 m
By using the concept of ray diagram of plane mirror shown below
Now, using the concept of similar triangle,
$\Delta$HAI $\sim$ $\Delta$GAE and $\Delta$BAI $\sim$ $\Delta$CAE
$\therefore$ ${{AI} \over {AE}} = {{HI} \over {EG}}$
$ \Rightarrow {{0.60} \over {1.8}} = {{0.25} \over {EG}}$ ($\because$ AI = IS)
$ \Rightarrow EG = 0.25 \times {{1.8} \over {0.6}} = 0.25 \times 3 = 0.75$ m
As, $CG = 2EG$
$ \Rightarrow CG = 0.75 \times 2 = 1.50$ m
Hence, distance between the extreme points, where he can see image of light source in mirror is 150 cm.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is __________ cm.
Correct Answer: 15
Explanation:
${1 \over v} - {1 \over u} = {1 \over f}$ .... (1)
m = ${v \over u}$ ..... (2)
from (1) and (2) we get
m = ${f \over {f + u}}$
given conditions
m1 = -m2
${f \over {f - 10}} = {{ - f} \over {f - 20}}$
$ \Rightarrow $ f – 20 = -f + 10
$ \Rightarrow $ 2f = 30
$ \Rightarrow $ f = 15 cm
m = ${v \over u}$ ..... (2)
from (1) and (2) we get
m = ${f \over {f + u}}$
given conditions
m1 = -m2
${f \over {f - 10}} = {{ - f} \over {f - 20}}$
$ \Rightarrow $ f – 20 = -f + 10
$ \Rightarrow $ 2f = 30
$ \Rightarrow $ f = 15 cm
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of
curvature of a surface of a plano-convex lens made of the same material with power 1.5 P is :
A.
${R \over 3}$
B.
${{3R} \over 2}$
C.
${R \over 2}$
D.
2R
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
A point like object is placed at a distance of 1 m in front of a convex lens of focal length 0.5 m. A
plane mirror is placed at a distance of 2 m behind the lens. The position and nature of the final
image formed by the system is :
A.
2.6 m from the mirror, real
B.
1 m from the mirror, real
C.
2.6 m from the mirror, virtual
D.
1 m from the mirror, virtual
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
For a concave lens of focal length f, the relation
between object and image distances u and v,
respectively, from its pole can best be
represented by (u = v is the reference line) :
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
A spherical mirror is obtained as shown in the
figure from a hollow glass sphere. If an object
is positioned in front of the mirror, what will be
the nature and magnification of the image of
the object? (Figure drawn as schematic and not to scale)
A.
Inverted, real and unmagnified
B.
Inverted, real and magnified
C.
Erect, virtual and magnified
D.
Erect, virtual and unmagnified
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
There is a small source of light at some depth
below the surface of water (refractive
index = ${4 \over 3}$) in a tank of large cross sectional
surface area. Neglecting any reflection from the
bottom and absorption by water, percentage of
light that emerges out of surface is (nearly) :
[Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2$\pi $rh]:
[Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2$\pi $rh]:
A.
17%
B.
34%
C.
50%
D.
21%
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
The aperture diameter of a telescope is 5m. The
separation between the moon and the earth is
4 × 105 km. With light of wavelength of
5500 $\mathop A\limits^o $, the minimum separation between
objects on the surface of moon, so that they are
just resolved, is close to :
A.
20 m
B.
200 m
C.
600 m
D.
60 m
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
A vessel of depth 2h is half filled with a liquid
of refractive index $2\sqrt 2 $ and the upper half with
another liquid of refractive index $\sqrt 2 $ . The
liquids are immiscible. The apparent depth of
the inner surface of the bottom of vessel
will be :
A.
${h \over {\sqrt 2 }}$
B.
${h \over {3\sqrt 2 }}$
C.
${3 \over 4}h\sqrt 2 $
D.
${h \over {2\left( {\sqrt 2 + 1} \right)}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
An object is gradually moving away from the
focal point of a concave mirror along the axis
of the mirror. The graphical representation of
the magnitude of linear magnification (m)
versus distance of the object from the mirror
(x) is correctly given by :
(Graphs are drawn schematically and are not to scale)
(Graphs are drawn schematically and are not to scale)
A.
B.
C.
D.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The magnifying power of a telescope with tube
60 cm is 5. What is the focal length of its eye
piece ?
A.
40 cm
B.
10 cm
C.
30 cm
D.
20 cm