Question:
A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface. A small air bubble is 6 cm below the flat surface inside it along the axis. The position of the image of the air bubble made by the mirror is seen :
A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface. A small air bubble is 6 cm below the flat surface inside it along the axis. The position of the image of the air bubble made by the mirror is seen :
Show Hint
The silvered curved surface is a concave mirror of radius 10 cm; locate the image by the mirror, then find its apparent depth through the flat surface (divide by 1.5).
Updated On: Nov 11, 2026
- 14 cm below flat surface
- 30 cm below flat surface
- 20 cm below flat surface
- 16 cm below flat surface
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The silvered curved surface is a concave mirror with \(R = 10\ \mathrm{cm}\), so \(f = -5\ \mathrm{cm}\). The centre of curvature is at the centre of the flat face.
The bubble is 6 cm below the flat surface, i.e. \(u = -(10 - 6) = -4\ \mathrm{cm}\) from the mirror pole.
\(\frac1v = \frac1f - \frac1u = -\frac15 + \frac14 = \frac1{20}\), so \(v = +20\ \mathrm{cm}\): a virtual image 20 cm behind the pole, i.e. \(10 + 20 = 30\ \mathrm{cm}\) below the flat surface (inside the glass medium).
This image is seen from outside through the flat surface, where the apparent depth is \(\frac{30}{1.5} = 20\ \mathrm{cm}\) below the flat surface.
Answer: (3) 20 cm below flat surface
The bubble is 6 cm below the flat surface, i.e. \(u = -(10 - 6) = -4\ \mathrm{cm}\) from the mirror pole.
\(\frac1v = \frac1f - \frac1u = -\frac15 + \frac14 = \frac1{20}\), so \(v = +20\ \mathrm{cm}\): a virtual image 20 cm behind the pole, i.e. \(10 + 20 = 30\ \mathrm{cm}\) below the flat surface (inside the glass medium).
This image is seen from outside through the flat surface, where the apparent depth is \(\frac{30}{1.5} = 20\ \mathrm{cm}\) below the flat surface.
Answer: (3) 20 cm below flat surface
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