Question:
A planoconvex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A.
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 :
A planoconvex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A.
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 :
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 :
Show Hint
Silvered system: Fig A: 2(n-1)/R = 1/28; Fig B (curved surface silvered): 2n/R = 1/10.
Updated On: Nov 11, 2026
- 1.50
- 1.55
- 1.75
- 1.51
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Let R be the radius of the curved surface.
Fig A (plane surface silvered): light refracts at the curved surface, reflects from the plane mirror (no power) and refracts again at the curved surface: \(\frac{1}{F_A} = \frac{2(n-1)}{R} = \frac{1}{28}\) ... (i)
Fig B (curved surface silvered, light enters through the plane face): the plane surface has no power; the curved mirror in glass has power \(\frac{2n}{R}\) (equivalently two passes through the lens, \(\frac{2(n-1)}{R}\), plus a mirror of power \(\frac2R\)): \(\frac{1}{F_B} = \frac{2n}{R} = \frac{1}{10}\) ... (ii)
Dividing (ii) by (i): \(\frac{n}{n-1} = \frac{28}{10} = 2.8\Rightarrow n = 2.8n - 2.8 \Rightarrow 1.8n = 2.8\Rightarrow n = 1.556 \approx 1.55\).
Answer: (2) 1.55
Fig A (plane surface silvered): light refracts at the curved surface, reflects from the plane mirror (no power) and refracts again at the curved surface: \(\frac{1}{F_A} = \frac{2(n-1)}{R} = \frac{1}{28}\) ... (i)
Fig B (curved surface silvered, light enters through the plane face): the plane surface has no power; the curved mirror in glass has power \(\frac{2n}{R}\) (equivalently two passes through the lens, \(\frac{2(n-1)}{R}\), plus a mirror of power \(\frac2R\)): \(\frac{1}{F_B} = \frac{2n}{R} = \frac{1}{10}\) ... (ii)
Dividing (ii) by (i): \(\frac{n}{n-1} = \frac{28}{10} = 2.8\Rightarrow n = 2.8n - 2.8 \Rightarrow 1.8n = 2.8\Rightarrow n = 1.556 \approx 1.55\).
Answer: (2) 1.55
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
Top JEE Main Ray optics and optical instruments Questions
- Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d < f1, d < f2) and their centers are displaced by a vertical separation Δ as shown in the figure.
- The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is ______ cm.
- If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be :
A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is:
- The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is __________ cm.
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions