Identify the correct statements.
If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig-B, it acts like an optical system of focal length $10$ $cm.$ The refractive index of the material of lens is :
| Object Pin |
Convex Lens |
Convex Mirror |
Image Pin |
|---|---|---|---|
| 22.2 cm | 32.2 cm | 45.8 cm | 71.2 cm |
The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to :
As the beam enters the medium, it will
The speed of light in the medium is
The incident angle $\theta $ for which the light ray grazes along the wall of the rod is :
A convex lens of refractive index 1.5 and focal length $f=18 \mathrm{~cm}$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times \mathrm{f}$. The value of $\alpha$ is $\_\_\_\_$ .
(refractive index of water $=4 / 3$ )
Explanation:
The Lens Maker's Formula is
$ \frac{1}{\mathrm{f}}=\left(\frac{\mu_1}{\mu_{\mathrm{m}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) $
where $\mu_1$ is the refractive index of the lens and $\mu_m$ is the refractive index of the surrounding medium.
Case-1 Focal length in Air $\left(\mathrm{f}_{\mathrm{a}}\right)$
In air, $\mu_{\mathrm{m}}=1$.
$ \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\frac{\mu_1}{1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) $
Let $\mathrm{K}=\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \Rightarrow \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\mu_1-1\right) \mathrm{K}$
Given $\mu_1=1.5$ and $\mathrm{f}_{\mathrm{a}}=\mathrm{f}=18 \mathrm{~cm}$ :
$ \frac{1}{\mathrm{f}}=(1.5-1) \mathrm{K}=0.5 \mathrm{~K} \Rightarrow \mathrm{~K}=\frac{2}{\mathrm{f}} $
Case-2 Focal length in Water ( $\mathrm{f}_{\mathrm{w}}$ )
In water, $\mu_{\mathrm{m}}=\mu_{\mathrm{w}}=4 / 3$ :
$\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{\mu_{\mathrm{l}}}{\mu_{\mathrm{w}}}-1\right) \mathrm{K}$
$\Rightarrow $ $\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{1.5}{4 / 3}-1\right) \mathrm{K}=\left(\frac{4.5}{4}-1\right) \mathrm{K}=\left(\frac{9}{8}-1\right) \mathrm{K}$
$ \frac{1}{\mathrm{f}_{\mathrm{w}}}=\frac{1}{8} \mathrm{~K}=\frac{1}{8} \times(2 \mathrm{f})=\frac{1}{4} \mathrm{f} $
Therefore, $\mathrm{f}_{\mathrm{w}}=4 \mathrm{f}$.
The difference in focal lengths is given as :
$ \Delta \mathrm{f}=\mathrm{f}_{\mathrm{w}}-\mathrm{f}_{\mathrm{a}} $
$\Rightarrow $ $ \Delta \mathrm{f}=4 \mathrm{f}-\mathrm{f}=3 \mathrm{f}=\alpha \times \mathrm{f} \Rightarrow \alpha=3 $
Therefore, the value of $\alpha$ is 3.
Hence, the correct answer is 3.
The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is $\_\_\_\_$ cm.
Explanation:
For a thin lens to form images of the same size at two different object positions, one image must be real and the other must be virtual. This typically occurs with a convex lens.
If the object distance is $u$, focal length of lens is $f$ then lens formula is given as,
$ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{u+f}{u f} $
The magnification is given as $m=\frac{v}{u} \Rightarrow m=\frac{1}{u} \times\left[\frac{u f}{u+f}\right]=\frac{f}{f+u}$
Where:
f is the focal length.
u is the object distance (using sign convention, u is negative).
Since the sizes (magnitudes of magnification) are equal but the nature of the images is different:
$ \left|\mathrm{m}_1\right|=\left|\mathrm{m}_2\right| \Rightarrow \mathrm{m}_1=-\mathrm{m}_2 $
Given:
1. Position $1=\mathrm{u}_1=-8 \mathrm{~cm}$
2. Position $2=\mathrm{u}_2=-24 \mathrm{~cm}$
$\frac{f}{f+(-8)}=-\left(\frac{f}{f+(-24)}\right)$
$\Rightarrow $ $\frac{1}{f-8}=-\frac{1}{f-24}$
$\Rightarrow $ $f-24=-(f-8)$
$\Rightarrow $ $f-24=-f+8$
$\Rightarrow $ $2 \mathrm{f}=32 $
$\Rightarrow $ $ \mathrm{f}=16 \mathrm{~cm}$
The focal length of the lens that produces images of equal size at object distances of 8 cm and 24 cm is 16 cm.
Hence, the correct answer is 16.
A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm . Refractive index of glass is 1.5 . The rays converge to a point at a distance $x \mathrm{~cm}$ from the centre of the curvature of the spherical surface. The value of $x$ is $\_\_\_\_$ cm .
Explanation:
Imagine a point object $O$ placed on the principal axis of a convex spherical surface that separates two media of refractive indices $\mu_1$ and $\mu_2$.
Here the pole of the surface be P and its centre of curvature be C.
Using the formula for refraction at a spherical surface :
$ \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} $
The incident beam is parallel, which means the light is coming from an object at infinity. Therefore, the object distance $u=-\infty$.
The refractive index of the first medium (air) is $\mu_1=1.0$.
The refractive index of the second medium (glass) is $\mu_2=1.5$.
The surface is convex facing the air. This means its centre of curvature lies inside the glass, in the direction of the traveling light. Thus, the radius of curvature is positive: $\mathrm{R}=+50 \mathrm{~cm}$.
Let's plug these into our derived equation to find the image position (v) :
$ \frac{1.5}{v}-\frac{1.0}{-\infty}=\frac{1.5-1.0}{+50} $
$\Rightarrow $ $\frac{1.5}{v}-0=\frac{0.5}{50}$
$\Rightarrow $ $\frac{1.5}{v}=\frac{1}{100}$
$\Rightarrow $ $ \mathrm{v}=150 \mathrm{~cm} $
So, the parallel rays will converge to form an image at a distance of 150 cm to the right of the pole of the spherical surface.
Pole $(P)$ is at 0 cm . and the centre of curvature $(C)$ is at +50 cm .
Image point / Convergence point (I) is at +150 cm .
So, the distance between C and I is
$ x=v-R $
$\Rightarrow $ $ \mathrm{x}=150 \mathrm{~cm}-50 \mathrm{~cm}=100 \mathrm{~cm} $
Therefore, the distance from the centre of curvature is 100 cm .
Hence, the correct answer is $\mathbf{1 0 0}$.
In a microscope the objective is having focal length $f_o=2 \mathrm{~cm}$ and eye-piece is having focal length $f_e=4 \mathrm{~cm}$. The tube length is 32 cm . The magnification produced by this microscope for normal adjustment is $\_\_\_\_$ .
Explanation:
In microscope normal adjustment refers to the condition where the final image is formed at infinity.
The total magnifying power ( M ) of a compound microscope in normal adjustment is the product of the linear magnification of the objective ( $\mathrm{m}_{\mathrm{o}}$ ) and the angular magnification of the eyepiece ( $\mathrm{m}_{\mathrm{e}}$ ):
$ \mathrm{M}=\mathrm{m}_{\mathrm{o}} \times \mathrm{m}_{\mathrm{e}} $
For an image at infinity:
Objective magnification $\mathrm{m}_{\mathrm{o}}=\frac{\mathrm{L}}{\mathrm{f}_{\mathrm{O}}}$, where L is the tube length.
Eyepiece magnification $\mathrm{m}_{\mathrm{e}}=\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}$, where D is the least distance of distinct vision (standard value is 25 cm ).
Thus, the total magnification formula is:
$ M=\frac{L}{f_o} \times \frac{D}{f_e} $
The focal length of objective is $\mathrm{f}_{\mathrm{o}}=2 \mathrm{~cm}$
The focal length of eyepiece is $\mathrm{f}_{\mathrm{e}}=4 \mathrm{~cm}$
The tube length is $\mathrm{L}=32 \mathrm{~cm}$
Substituting the values,
$ M=\frac{32}{2} \times \frac{25}{4} 100 $
Therefore, the total magnification of given microscope is 100 . Hence, the correct answer is $\mathbf{1 0 0}$.
A collimated beam of light of diameter 2 mm is propagating along $x$-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm , then the focal length of second lens is $\_\_\_\_$ mm.
Explanation:
For a beam to enter collimated (parallel rays) and exit collimated with a larger diameter, the two lenses must be placed such that their focal points coincide.
The first lens focuses the parallel rays at its focal point ( $f_1$ ) and then the second lens is placed so that its focal point $\left(f_2\right)$ is at the same location. It then re-collimates the rays diverging from that point.
From similar triangles in the ray diagram, the ratio of the diameters of the beams is directly proportional to the ratio of the focal lengths of the lenses.
$ \frac{D_2}{D_1}=\frac{f_2}{f_1} $
Where :
$\mathrm{D}_1=$ Initial beam diameter $=2 \mathrm{~mm}$
$\mathrm{D}_2=$ Final beam diameter $=14 \mathrm{~mm}$
$\mathrm{f}_1=$ Focal length of the first lens $=40 \mathrm{~mm}$
$f_2=$ Focal length of the second lens (need to be calculated)
$\Rightarrow $$\mathrm{f}_2=\mathrm{f}_1 \times\left(\frac{\mathrm{D}_2}{\mathrm{D}_1}\right)$
$\Rightarrow $ $\mathrm{f}_2=40 \times\left(\frac{14}{2}\right)$
$\Rightarrow $ $ \mathrm{f}_2=280 \mathrm{~mm} $
Therefore, to expand the beam from a diameter of 2 mm to 14 mm using a first lens of 40 mm focal length, the second lens must have a focal length of 280 mm .
Hence, the correct answer is 280 .
A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40 cm . The value of H is ________ cm . (Consider liquids are immisible)
Explanation:
$\begin{aligned} & y=\text { apparent depth of bottom } \\ & \frac{y}{1}=\frac{H}{1.6}+\frac{60}{1.2} \\ & \text { Shift }=40 \\ & H+60-y=40 \\ & H+60-\frac{\mathrm{H}}{1.6}-\frac{60}{1.2}=40 \\ & \frac{6}{16} \mathrm{H}=30 \\ & H=80 \mathrm{~cm} \end{aligned}$
Distance between object and its image (magnified by $-\frac{1}{3}$ ) is 30 cm . The focal length of the mirror used is $\left(\frac{x}{4}\right) \mathrm{cm}$, where magnitude of value of $x$ is _________.
Explanation:
$\begin{aligned} &\begin{aligned} & M=-\frac{1}{3} \\ & -\frac{-V}{-U}=\frac{-1}{3} \Rightarrow V=\frac{U}{3} \end{aligned}\\ &\text { Distance b/w object and image : } \end{aligned}$
$\begin{aligned} & U-V=30 \\ & U-\frac{\mu}{3}=30 \\ & \Rightarrow U=45 \quad V=15 \\ & \frac{1}{f}=\frac{1}{V}+\frac{1}{U}=-\frac{1}{15}-\frac{1}{45} \\ & \Rightarrow F=\frac{45}{4} \\ & x=45 \end{aligned}$
Light from a point source in air falls on a spherical glass surface (refractive index, $\mu=1.5$ and radius of curvature $=50 \mathrm{~cm}$ ). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ___________m.
Explanation:
We apply the refraction formula at a spherical surface: μ₂/v − μ₁/u = (μ₂ − μ₁)/R.
Here:
• μ₁ = 1 (air)
• μ₂ = 1.5 (glass)
• R = 50 cm (radius of curvature)
• v = +200 cm (image distance inside glass)
• u = –x (object distance, negative for a real object in front of the surface)
Substitute these values and solve for x:
$\begin{aligned} & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{200}-\frac{1}{-\mathrm{x}}=\frac{1.5-1}{50} \\ & \frac{1}{\mathrm{x}}=\frac{1}{100}-\frac{3}{400} \\ & \mathrm{x}=400 \mathrm{~cm} \\ & \mathrm{x}=4 \mathrm{~m} \end{aligned}$
So, the light source is 4 m from the glass surface.
Explanation:
To find the angle of incidence when a ray of light experiences minimum deviation in a prism, we use the formula for the refractive index $\mu$:
$ \mu = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} $
Given:
The angle of the prism, $A = 60^{\circ}$
The refractive index of the prism material, $\mu = \sqrt{2}$
Minimum deviation, $\delta_m = 30^{\circ}$
At minimum deviation, the angle of incidence $i$ equals the angle of emergence $e$. Therefore, the relation between the angle of minimum deviation and the angle of the prism is represented as:
$ \delta_m = 2i - A $
Solving for $i$:
$ \delta_m = 2i - A \implies 30^{\circ} = 2i - 60^{\circ} \implies 2i = 90^{\circ} \implies i = 45^{\circ} $
Thus, the angle of incidence is $45^{\circ}$.