Geometrical Optics
The distance between object and its two times magnified real image as produced by a convex lens is $45 \mathrm{~cm}$. The focal length of the lens used is _______ cm.
Explanation:
$\begin{aligned} & \frac{v}{u}=-2 \\ & v=-2 u \quad\text{... (i)} \\ & v-u=45 \quad\text{... (ii)}\\ & \Rightarrow u=-15 \mathrm{~cm} \\ & v=30 \mathrm{~cm} \\ & \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ & f=+10 \mathrm{~cm} \end{aligned}$
Two immiscible liquids of refractive indices $\frac{8}{5}$ and $\frac{3}{2}$ respectively are put in a beaker as shown in the figure. The height of each column is $6 \mathrm{~cm}$. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\frac{\alpha}{4} \mathrm{~cm}$. The value of $\alpha$ is _________.
Explanation:
$h_{\text {app }}=\frac{h_1}{\mu_1}+\frac{h_2}{\mu_2}=\frac{6}{3 / 2}+\frac{6}{8 / 5}=4+\frac{15}{4}=\frac{31}{4} \mathrm{~cm}$
A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The radius of curvature of the convex surface (SPU) is $9 \mathrm{~cm}$, while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index $n$ up to the level QPR. If the image of a point object $\mathrm{O}$ at a height of $h$ (OT in the figure) is formed onto itself, then, which of the following option(s) is(are) correct?
Two equilateral-triangular prisms $\mathrm{P}_1$ and $\mathrm{P}_2$ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism $\mathrm{P}_1$ at an angle of incidence $\theta$ such that the outgoing ray undergoes minimum deviation in prism $\mathrm{P}_2$. If the respective refractive indices of $\mathrm{P}_1$ and $\mathrm{P}_2$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$, then $\theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{\beta}\right)\right]$, where the value of $\beta$ is ____.
Explanation:
At surface BC
$\begin{aligned} & \sqrt{\frac{3}{2}} \sin r_2=\sqrt{3} \sin 30 \\ & \sqrt{\frac{3}{2}} \sin r_2=\frac{\sqrt{3}}{2} \\ & \operatorname{sinr}_2=\frac{1}{\sqrt{2}} \\ & r_2=45^{\circ} \\ & r_1=60^{\circ}-45^{\circ}=15^{\circ} \end{aligned}$
At surface $A B$
$\begin{gathered} 1 \sin \theta=\sqrt{\frac{3}{2}} \sin 15^{\circ} \\ \theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \frac{\pi}{12}\right] \\ \beta=12 \end{gathered}$
A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $\theta_0$. The ray partially refracts into the sphere with angle of refraction $\phi_0$ and then partly reflects from the back surface. The reflected ray then emerges out of the sphere after a partial refraction. The total angle of deviation of the emergent ray with respect to the incident ray is $\alpha$. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
| List-I | List-II |
|---|---|
| (P) If $n = 2$ and $\alpha = 180^\circ$, then all the possible values of $\theta_0$ will be | (1) $30^\circ$ and $0^\circ$ |
| (Q) If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $\theta_0$ will be | (2) $60^\circ$ and $0^\circ$ |
| (R) If $n = \sqrt{3}$ and $\alpha = 180^\circ$, then all the possible values of $\phi_0$ will be | (3) $45^\circ$ and $0^\circ$ |
| (S) If $n = \sqrt{2}$ and $\theta_0 = 45^\circ$, then all the possible values of $\alpha$ will be | (4) $150^\circ$ |
| (5) $0^\circ$ |
A vessel of depth '$d$' is half filled with oil of refractive index $n_{1}$ and the other half is filled with water of refractive index $n_{2}$. The apparent depth of this vessel when viewed from above will be-
An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is $12 \mathrm{~cm}$. When viewed from the opposite side, the apparent distance of the bubble is observed as $4 \mathrm{~cm}$. If the side of the ice cube is $24 \mathrm{~cm}$, the refractive index of the ice cube is
When one light ray is reflected from a plane mirror with $30^{\circ}$ angle of reflection, the angle of deviation of the ray after reflection is :
The critical angle for a denser-rarer interface is $45^{\circ}$. The speed of light in rarer medium is $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. The speed of light in the denser medium is:
An object is placed at a distance of 12 cm in front of a plane mirror. The virtual and erect image is formed by the mirror. Now the mirror is moved by 4 cm towards the stationary object. The distance by which the position of image would be shifted, will be
In a reflecting telescope, a secondary mirror is used to:
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: The phase difference of two light waves change if they travel through different media having same thickness, but different indices of refraction.
Reason R: The wavelengths of waves are different in different media.
In the light of the above statements, choose the most appropriate answer from the options given below
A 2 meter long scale with least count of $0.2 \mathrm{~cm}$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $80 \mathrm{~cm}$ mark and $1 \mathrm{~m}$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $180 \mathrm{~cm}$ mark. The $\%$ error in the estimation of focal length is:
A monochromatic light wave with wavelength $\lambda_{1}$ and frequency $v_{1}$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are $45^{\circ}$ and $30^{\circ}$ respectively, then the wavelength $\lambda_{2}$ and frequency $v_{2}$ of the refracted wave are:
Two objects A and B are placed at 15 cm and 25 cm from the pole in front of a concave mirror having radius of curvature 40 cm. The distance between images formed by the mirror is _______________.
A thin prism $P_1$ with an angle $6^{\circ}$ and made of glass of refractive index $1.54$ is combined with another prism $P_2$ made from glass of refractive index $1.72$ to produce dispersion without average deviation. The angle of prism $P_2$ is
A person has been using spectacles of power $-1.0$ dioptre for distant vision and a separate reading glass of power $2.0$ dioptres. What is the least distance of distinct vision for this person :
A scientist is observing a bacteria through a compound microscope. For better analysis and to improve its resolving power he should. (Select the best option)
The light rays from an object have been reflected towards an observer from a standard flat mirror, the image observed by the observer are :-
A. Real
B. Erect
C. Smaller in size then object
D. Laterally inverted
Choose the most appropriate answer from the options given below :
When a beam of white light is allowed to pass through convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called :
Explanation:
$n = \frac{\sin \frac{A + \delta_m}{2}}{\sin \frac{A}{2}}$
In this case, the refractive index $n = \sqrt{2}$, and since the prism is equilateral, the prism angle $A = 60^\circ$. Now, we can substitute these values into the formula and solve for the angle of minimum deviation $\delta_m$:
$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\sin \frac{60^\circ}{2}}$
First, let's find the sine of half the prism angle:
$\sin \frac{60^\circ}{2} = \sin 30^\circ = \frac{1}{2}$
Now, we can substitute this value into the formula:
$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\frac{1}{2}}$
To isolate the sine term, we multiply both sides by $\frac{1}{2}$:
$\frac{\sqrt{2}}{2} = \sin \frac{60^\circ + \delta_m}{2}$
Now, we can find the angle inside the sine function:
$\frac{60^\circ + \delta_m}{2} = \sin^{-1} \frac{\sqrt{2}}{2} = 45^\circ$
Finally, we can solve for the angle of minimum deviation $\delta_m$:
$60^\circ + \delta_m = 2 \cdot 45^\circ$
$\delta_m = 90^\circ - 60^\circ = 30^\circ$
The angle of minimum deviation for the liquid in the equilateral hollow prism is $30^\circ$.
A bi convex lens of focal length $10 \mathrm{~cm}$ is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is ____________ D.
Explanation:
$\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
For the original biconvex lens, both radii of curvature have the same magnitude but opposite signs, so let's denote them as ±R. The focal length of the original lens is given as 10 cm, and the refractive index (n) is constant for both the original lens and the new plano-convex lenses.
For the original lens, the lensmaker's formula becomes:
$\frac{1}{f} = (n - 1) \left(\frac{1}{R} - \frac{1}{-R}\right)$
Since the focal length is given as 10 cm, we can plug in the value:
$\frac{1}{10} = (n - 1) \left(\frac{1}{R} + \frac{1}{R}\right)$
$\frac{1}{10} = (n - 1) \left(\frac{2}{R}\right)$
For each plano-convex lens, one radius of curvature (R1) is infinite (the flat side), and the other radius (R2) is the same as the original lens (R). The lensmaker's formula for the plano-convex lens becomes:
$\frac{1}{f'} = (n - 1) \left(\frac{1}{\infty} - \frac{1}{R}\right)$
$\frac{1}{f'} = (n - 1) \left(-\frac{1}{R}\right)$
Now, we can substitute the expression for (n-1)(2/R) from the original lens equation:
$\frac{1}{f'} = \frac{1}{10} \cdot \frac{1}{2}$ $\frac{1}{f'} = \frac{1}{20}$
So, the focal length of each plano-convex lens (f') is 20 cm. To find the power of each lens, we can use the formula:
$P (\text{in diopters}) = \frac{1}{f (\text{in meters})}$
Converting the focal length to meters and calculating the power:
$P = \frac{1}{0.2}$
$P = 5 \text{ D}$
So, the power of each plano-convex lens after the cut is 5 diopters (D).
A fish rising vertically upward with a uniform velocity of $8 \mathrm{~ms}^{-1}$, observes that a bird is diving vertically downward towards the fish with the velocity of $12 \mathrm{~ms}^{-1}$. If the refractive index of water is $\frac{4}{3}$, then the actual velocity of the diving bird to pick the fish, will be __________ $\mathrm{ms}^{-1}$.
Explanation:
The fish sees the bird diving with a velocity of $12 \ \text{m/s}$. We can write the equation considering the velocities relative to the fish:
$\frac{V_{\text{b/f}}}{\frac{4}{3}} = \frac{-8}{\frac{4}{3}} + \frac{-v}{1}$
Here, $V_{\text{b/f}}$ is the bird's diving velocity relative to the fish, and $v$ is the actual velocity of the bird.
Now, let's solve for $v$:
$\frac{-12}{\frac{4}{3}} = \frac{-8}{\frac{4}{3}} - v$
$v = 3 \ \text{m/s}$
So, the actual velocity of the diving bird to pick the fish relative to the fish is $3 \ \text{m/s}$.
Two convex lenses of focal length $20 \mathrm{~cm}$ each are placed coaxially with a separation of $60 \mathrm{~cm}$ between them. The image of the distant object formed by the combination is at _____________ $\mathrm{cm}$ from the first lens.
Explanation:
1. First refraction in L1 (lens 1):
When considering the first lens (L1), the object is at infinity, so the image I1 formed by this lens is at its focal point. Using the lensmaker's equation:
$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$
Given that the object is at infinity, $d_o = \infty$, and the focal length of the first lens is $f = 20 \mathrm{~cm}$. Plugging in these values, we get:
$\frac{1}{20} = \frac{1}{\infty} + \frac{1}{d_i}$
As $\frac{1}{\infty}$ is essentially 0, we have:
$\frac{1}{20} = \frac{1}{d_i}$
This implies that $d_i = 20 \mathrm{~cm}$, meaning that the image I1 is formed 20 cm from the first lens L1.
2. Second refraction in L2 (lens 2):
Now, the image I1 formed by L1 becomes the object for L2. The distance between the lenses is 60 cm, so the object distance for L2 (u) is -40 cm, because the object is to the left of the lens (u is negative). The focal length of L2 (f) is 20 cm. We can use the lensmaker's equation to find the image distance (v) for L2:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Plugging in the values:
$\frac{1}{v} - \frac{1}{(-40)} = \frac{1}{20}$
$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40}$
$\frac{1}{v} = \frac{1}{40}$
Therefore, $v = 40 \mathrm{~cm}$
Since the image distance (v) is positive, the image I2 is formed on the right side of L2 at a distance of 40 cm. To find the distance of the final image from L1, we add the distance between the lenses (60 cm) and the image distance (v) from L2.
Final image distance from L1 = 60 cm + 40 cm = 100 cm
Thus, the correct answer is 100 cm.
As shown in the figure, a plane mirror is fixed at a height of $50 \mathrm{~cm}$ from the bottom of tank containing water $\left(\mu=\frac{4}{3}\right)$. The height of water in the tank is $8 \mathrm{~cm}$. A small bulb is placed at the bottom of the water tank. The distance of image of the bulb formed by mirror from the bottom of the tank is ___________ $\mathrm{cm}$.
Explanation:
$ \text { Apparent depth of } \mathrm{O}=\frac{\mathrm{d}}{\mu} $$ =\frac{8}{\frac{4}{3}}=6 \mathrm{~cm} $
Distance of object from mirror = 42 + 6 = 48 m
$ \text { Distance between } \mathrm{O} \text { and } \mathrm{I}_2=48+50=98 \mathrm{~cm} $
The radius of curvature of each surface of a convex lens having refractive index 1.8 is $20 \mathrm{~cm}$. The lens is now immersed in a liquid of refractive index 1.5 . The ratio of power of lens in air to its power in the liquid will be $x: 1$. The value of $x$ is _________.
Explanation:
Let's find the focal length of the lens in air and in the liquid. We will use the lens maker's formula:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
where $f$ is the focal length, $\mu$ is the refractive index of the lens material, and $R_1$ and $R_2$ are the radii of curvature of the lens surfaces.
Since the lens is convex, both surfaces have the same radius of curvature (positive), so $R_1 = R_2 = 20 \mathrm{~cm}$.
First, let's find the focal length of the lens in air:
$\frac{1}{f_\text{air}} = (1.8 - 1)\left(\frac{1}{20} - \frac{1}{20}\right) = 0.8\left(\frac{1}{20}\right)$
$f_\text{air} = \frac{1}{0.8\left(\frac{1}{20}\right)} = 25 \mathrm{~cm}$
Now, let's find the focal length of the lens in the liquid. The relative refractive index of the lens with respect to the liquid is:
$\mu_\text{rel} = \frac{1.8}{1.5} = 1.2$
$\frac{1}{f_\text{liquid}} = (1.2 - 1)\left(\frac{1}{20}\right)$
$f_\text{liquid} = \frac{1}{0.2\left(\frac{1}{20}\right)} = 100 \mathrm{~cm}$
The power of a lens is given by:
$P = \frac{1}{f}$
Now, we can find the ratio of the power of the lens in air to its power in the liquid:
$\frac{P\text{air}}{P\text{liquid}} = \frac{f\text{liquid}}{f\text{air}} = \frac{100}{25} = 4$
So, the ratio of the power of the lens in air to its power in the liquid is $x:1$, where $x = 4$.
A point object, 'O' is placed in front of two thin symmetrical coaxial convex lenses $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ with focal length $24 \mathrm{~cm}$ and $9 \mathrm{~cm}$ respectively. The distance between two lenses is $10 \mathrm{~cm}$ and the object is placed $6 \mathrm{~cm}$ away from lens $\mathrm{L}_{1}$ as shown in the figure. The distance between the object and the image formed by the system of two lenses is __________ $\mathrm{cm}$.
Explanation:
$\frac{1}{\mathrm{v}}+\frac{1}{6}=\frac{1}{24}$
$ \begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{24}-\frac{1}{6}=-\frac{1}{8} \\\\ \mathrm{v} & =-8 \mathrm{~cm} \end{aligned} $
From IInd lens :
$\frac{1}{\mathrm{v}}+\frac{1}{18}=\frac{1}{9}$
$ \begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{18} \\\\ \mathrm{v} & =18 \end{aligned} $
So distance between object and its image
= 6 + 10 + 18 = 34 cm
Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature $30 \mathrm{~cm}$. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of $15 \mathrm{~cm}$ from the pole of the surface. The distance of image from the pole of the surface is ____________ $\mathrm{cm}$.
Explanation:
The refraction at a spherical surface is governed by the formula:
$\frac{1}{v} - \frac{1}{u} = \frac{n_2 - n_1}{R} n_2$
where:
- (v) is the image distance,
- (u) is the object distance,
- ($n_1$) is the refractive index of the medium where the object is,
- ($n_2$) is the refractive index of the medium where the image is formed,
- (R) is the radius of curvature of the refracting surface.
Here, we have:
- (u = -15) cm (the object is on the same side as the light is coming from, so the distance is taken as negative),
- ($n_1$ = 1.0) (refractive index of the rarer medium),
- ($n_2$ = 1.5) (refractive index of the denser medium),
- (R = -30) cm (the center of curvature is in the denser medium, so the radius is taken as negative).
Substituting these values into the formula, we get:
$\frac{1}{v} - \left(-\frac{1}{15}\right) = \frac{1.5 - 1.0}{-30} \cdot 1.5$
Solving for (v), we get:
$v = \frac{1}{\frac{1}{15} + \frac{1.5}{30}} = -30 \, \text{cm}$
Therefore, the image is formed at a distance of 30 cm from the pole of the surface, on the same side as the object. The negative sign indicates that the image is virtual and is formed on the same side of the surface as the light is coming from.
Two vertical parallel mirrors A and B are separated by $10 \mathrm{~cm}$. A point object $\mathrm{O}$ is placed at a distance of $2 \mathrm{~cm}$ from mirror $\mathrm{A}$. The distance of the second nearest image behind mirror A from the mirror $\mathrm{A}$ is _________ $\mathrm{cm}$.
Explanation:
Object distance $=8 \mathrm{~cm}$ [For image by $B]$
$ \begin{aligned} \Rightarrow \text { Required distance } & =8+8+2 \mathrm{~cm} \\\\ & =18 \mathrm{~cm} \end{aligned} $
A pole is vertically submerged in swimming pool, such that it gives a length of shadow $2.15 \mathrm{~m}$ within water when sunlight is incident at angle of $30^{\circ}$ with the surface of water. If swimming pool is filled to a height of $1.5 \mathrm{~m}$, then the height of the pole above the water surface in centimeters is $\left(n_{w}=4 / 3\right)$ ____________.
Explanation:
The pole is vertically submerged in the swimming pool, and the length of the shadow is due to the sunlight that is incident at an angle of $30^{\circ}$ with the surface of water. Hence, the angle of incidence, $i$, is $60^{\circ}$ (since the angle of incidence is measured from the normal to the surface, and the normal is perpendicular to the surface).
Using Snell's law, we find the angle of refraction, $r$, as you correctly did:
$\sin r = \frac{n_{1}}{n_{2}} \sin i = \frac{3}{4} \sin 60^{\circ} = \frac{3 \sqrt{3}}{8}$
This gives $\tan r = \frac{3 \sqrt{3}}{\sqrt{37}}$.
Now, the shadow of the pole in the water forms a right triangle, with the submerged part of the pole as one side, the shadow as the hypotenuse, and the line segment from the water surface to the end of the shadow as the other side. The angle at the water surface is $r$. Thus, we can write the following relationships:
The length of the submerged part of the pole (which I'll call $x$), is given by:
$x = 2.15 \, \text{m} \cdot \cos r$
And the length of the line segment from the water surface to the end of the shadow (which I'll call $y$), is given by:
$y = 2.15 \, \text{m} \cdot \sin r$
The total length of the pole above the water surface is then the sum of $x$ and the depth of the swimming pool (1.5 m), which gives us the equation:
$x \sqrt{3} + 1.5 \, \text{m} \cdot \tan r = 2.15 \, \text{m}$
Solving this for $x$ gives us:
$x = \frac{2.15 \, \text{m}}{\sqrt{3}} - \frac{1.5 \, \text{m} \cdot 3}{\sqrt{37}} = 0.502 \, \text{m} = 50.2 \, \text{cm}$
So, the height of the pole above the water surface is approximately 50 cm.
A thin cylindrical rod of length $10 \mathrm{~cm}$ is placed horizontally on the principle axis of a concave mirror of focal length $20 \mathrm{~cm}$. The rod is placed in a such a way that mid point of the rod is at $40 \mathrm{~cm}$ from the pole of mirror. The length of the image formed by the mirror will be $\frac{x}{3} \mathrm{~cm}$. The value of $x$ is _____________.
Explanation:
$\text { A: }$
$ \begin{aligned} & \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\\\ & \Rightarrow \frac{1}{v}+\frac{1}{-45}=\frac{1}{-20} \\\\ & \Rightarrow \frac{1}{v}=\frac{1}{45}-\frac{1}{20}=\frac{4-9}{180}=-\frac{1}{36} \\\\ & \Rightarrow v=-36 \mathrm{~cm} \end{aligned} $
B: $\frac{1}{v}+\frac{1}{-35}=\frac{1}{-20}$
$\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{20}=\frac{4-7}{140}$
$\Rightarrow \quad v=-\frac{140}{3}$
$\Rightarrow$ length of image $=\frac{140}{3}-36=\frac{32}{3} \mathrm{~cm}$
$\Rightarrow x=32$
In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $40 \mathrm{~cm}$ from the pole of the mirror is formed at distance $120 \mathrm{~cm}$ from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in $1 \mathrm{~cm}$. The value of error in measurement of focal length of the mirror is $\frac{1}{\mathrm{~K}} \mathrm{~cm}$. The value of $\mathrm{K}$ is __________.
Explanation:
${1 \over v} + {1 \over u} = {1 \over f}$ ...... (1)
$ \Rightarrow - {1 \over {{f^2}}}df = - {1 \over {{v^2}}}dv - {1 \over {{u^2}}}du$
$ \Rightarrow {{df} \over {{f^2}}} = {{dv} \over {{v^2}}} + {{du} \over {{u^2}}}$ ..... (2)
From (1) : $ - {1 \over {120}} - {1 \over {40}} = {1 \over f} \Rightarrow f = - 30$ cm
Also, least count $ = {{1\,\mathrm{cm}} \over {20}} = 0.05$ cm
$ \Rightarrow df = \left[ {{{0.05} \over {{{120}^2}}} + {{0.05} \over {{{40}^2}}}} \right] \times {30^2}$
$ = 0.05\left[ {{1 \over {16}} + {9 \over {16}}} \right] = {5 \over 8} \times {5 \over {100}} = {1 \over {32}}$ cm
$ \Rightarrow k = 32$
In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm and apparent thickness of the glass slab as 5.00 mm. Travelling microscope has 20 divisions in one cm on main scale and 20 divisions on vernier scale is equal to 49 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is $\frac{x}{10}\times10^{-3}$, where $x$ is ___________
Explanation:
$\mu=\frac{\mathrm{real\,depth}\,(l_1)}{\mathrm{apparent\,depth}\,(l_2)}$
$=\frac{5.25}{5}=1.05$
$\frac{d\mu}{\mu}=\frac{dl_1}{l_1}+\frac{dl_2}{l_2}$
$d\mu = \left( {{{d{l_1}} \over {{l_1}}} + {{d{l_2}} \over {{l_2}}}} \right)\mu $
$ = \left( {{{0.01} \over {5.25}} + {{0.01} \over {5.00}}} \right) \times 1.05$
$ = {{41} \over {10}} \times {10^{ - 3}}$
so $x = 41$
An object is placed on the principal axis of convex lens of focal length 10cm as shown. A plane mirror is placed on the other side of lens at a distance of 20 cm. The image produced by the plane mirror is 5cm inside the mirror. The distance of the object from the lens is ___________ cm.
Explanation:
$I_{1}$ is image formed by lens and $I_{2}$ is image formed by mirror.
Location of $I_{1}$ and $I_{2}$ from mirror will be equal $=5 \mathrm{~cm}$ Hence $\mathrm{I}_{1}=15 \mathrm{~cm}$ from lens
From $\frac{1}{v}-\frac{1}{u}=\frac{1}{10}$
$ u=-x, v=15 $
$\frac{1}{x}=\frac{1}{10}-\frac{1}{15} \Rightarrow x=30 \mathrm{~cm}$
A ray of light is incident from air on a glass plate having thickness $\sqrt3$ cm and refractive index $\sqrt2$. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is ____________ $\times$ 10$^{-2}$ cm. (given $\sin 15^\circ = 0.26$)
Explanation:
$\Rightarrow$ at point (1)
$\mu \sin r=\sin i=\frac{1}{\sqrt{2}}$
$\sin r=\frac{1}{2} \quad \Rightarrow \quad r=30^{\circ}$
Lateral displacement
$=\frac{t}{\cos r} \sin \left(15^{\circ}\right)=\frac{\sqrt{3}}{\left(\frac{\sqrt{3}}{2}\right)} \times 0.26$
$=2 \times 0.26$
$=0.52 \mathrm{~cm}$ $=52 \times 10^{-2} \mathrm{~cm}$
A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change in focal length of the lens will be ___________ cm.
(Given refractive index of water $=\frac{4}{3}$)
Explanation:
$ \frac{1}{f}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {mrdium }}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] $
when in air
$ \frac{1}{18}=\left(\frac{1.5}{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(1) $
$\mu_{\text {lense }}=1.5, \mu_{\text {air }}=1$.
when in water
$ \frac{1}{f}=\left(\frac{1.5}{4 / 3}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(2) $
from (1) & (2)
$f=72$
Change in focal length $=72-18 = 54$
As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is 30 cm and refraction index of the material for both the lenses is 1.75. Both the lenses are placed at distance of 40 cm from each other. Due to the combination, the image of the object is formed at distance $x=$ _________ cm, from concave lens.
Explanation:
$f_{\text {concave }}=-40 \mathrm{~cm}$
$\frac{1}{f_{\text {convex }}}=(1.75-1)\left(\frac{1}{30}-\frac{1}{\infty}\right)=\frac{0.75}{30}$
$f_{\text {convex }}=40 \mathrm{~cm}$
Let the first image is formed at $v_{1}$ so
$\frac{1}{v_{1}}-\frac{1}{\infty}=\frac{1}{f_{\text {concave }}}=-\frac{1}{40}$
$\Rightarrow v_{1}=-40 \mathrm{~cm}$
for second image
$\frac{1}{x-40}-\frac{1}{-80}=\frac{1}{40}$
$\Rightarrow x=120 \mathrm{~cm}$
Explanation:
Given that the object is positioned at the focal point, the resulting light rays will transition to a parallel alignment.
Upon the first reflection at Mirror $ M_1 $:
As the incoming light rays are parallel, the resultant image will be formed at the focal point.
$\mathrm{II}^{\text {nd }}$ refraction from $\mathrm{L}$ :-
$ \begin{aligned} & \mathrm{u}=-(\mathrm{d}-10) \\\\ & \mathrm{f}=10 \mathrm{~cm} \\\\ & \frac{1}{\mathrm{v}}-\frac{1}{\mu}=\frac{1}{\mathrm{f}} \\\\ & \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{~d}-10}=\frac{1}{10} \\\\ & \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{(\mathrm{~d}-10)} .......(i) \end{aligned} $
This $v$ will be object for $M_2$, and image should be at $10 \mathrm{~cm}$
$ \begin{aligned} & \frac{1}{\mu}+\frac{1}{v_1}=\frac{1}{f} \\\\ & -\frac{1}{(20-v)}-\frac{1}{10}=-\frac{1}{12} \\\\ & \frac{1}{12}-\frac{1}{10}=\frac{1}{20-v} \\\\ & -\frac{2}{120}=\frac{1}{20-v} \\\\ & 20-v=-60 \\\\ & v=80 \mathrm{~cm} \end{aligned} $
From equation (i), we get
$ \begin{aligned} & \frac{1}{80}=\frac{1}{10}-\frac{1}{d-10} \\\\ & \frac{1}{d-10}=\frac{1}{10}-\frac{1}{80} \\\\ & \frac{1}{d-10}=\frac{80-10}{800}=\frac{70}{800} \\\\ & d-10=\frac{80}{7} \Rightarrow d=10+\frac{80}{7}=\frac{150}{7} \\\\ & n=150 \end{aligned} $
Case-2 : If $1^{\text {st }}$ reflection on mirror $m_2$ :
For $\mathrm{m}_2$
$ \begin{aligned} & \frac{1}{\mathrm{~V}_1}+\frac{1}{-10}=\frac{1}{-12} \\\\ & \mathrm{~V}_1=60 \mathrm{~cm} \end{aligned} $
Then refraction on lens $\mathrm{L}$
$ \begin{aligned} & \mathrm{u}_2=80 \mathrm{~cm} \\\\ & \frac{1}{\mathrm{~V}_2}-\frac{1}{-60}=\frac{1}{10} \\\\ & \mathrm{~V}_2=\frac{80}{7} \end{aligned} $
Then reflection on $\mathrm{m}_2$
Either $V_2$ is at centre (normal incidence)
$ \begin{aligned} & \mathrm{d}-\frac{80}{7}=20 \\\\ & \mathrm{~d}=\frac{220}{7} \\\\ & \frac{\mathrm{n}}{7}=\frac{220}{7} \\\\ & \mathrm{n}=220 \end{aligned} $
$V_2$ is at pole of $m_2$
$ \begin{aligned} & \mathrm{d}-\frac{80}{7}=0 \\\\ & \mathrm{~d}=\frac{80}{7} \\\\ & \frac{\mathrm{n}}{7}=\frac{80}{7} \\\\ & \mathrm{n}=80 \end{aligned} $
Light enters from air into a given medium at an angle of $45^{\circ}$ with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of $15^{\circ}$ from its original direction. The refractive index of the medium is:
The power of a lens (biconvex) is $1.25 \mathrm{~m}^{-1}$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are $20 \mathrm{~cm}$ and $40 \mathrm{~cm}$ respectively. The refractive index of surrounding medium:
As shown in the figure, after passing through the medium 1 . The speed of light $v_{2}$ in medium 2 will be :
$\left(\right.$ Given $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )
In normal adujstment, for a refracting telescope, the distance between objective and eye piece is $30 \mathrm{~cm}$. The focal length of the objective, when the angular magnification of the telescope is 2 , will be :
A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $\lambda$, calculate the change of microscope's resolving power due to oil and choose the correct option.
Light travels in two media $M_{1}$ and $M_{2}$ with speeds $1.5 \times 10^{8} \mathrm{~ms}^{-1}$ and $2.0 \times 10^{8} \mathrm{~ms}^{-1}$ respectively. The critical angle between them is :
For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?
Which of the following statement is correct?
Time taken by light to travel in two different materials $A$ and $B$ of refractive indices $\mu_{A}$ and $\mu_{B}$ of same thickness is $t_{1}$ and $t_{2}$ respectively. If $t_{2}-t_{1}=5 \times 10^{-10}$ s and the ratio of $\mu_{A}$ to $\mu_{B}$ is $1: 2$. Then, the thickness of material, in meter is: (Given $v_{\mathrm{A}}$ and $v_{\mathrm{B}}$ are velocities of light in $A$ and $B$ materials respectively.)