A Hemispherical Portion of the Surface of a Solid Glass Sphere (μ = 1.5) of Radius R is Silvered to Make the Inner Side Reflecting. an Object is Placed on the Axis of the Hemisphere at - Physics

Sum

A hemispherical portion of the surface of a solid glass sphere (μ = 1.5) of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.

Solution

As shown in the figure, OQ = 3r and OP = r
Thus, PQ = 2r
For refraction at APB,
we know that,

$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$

$\Rightarrow \frac{1 . 5}{v} - \left( \frac{1}{- 2r} \right) = \frac{0 . 5}{r} = \frac{1}{2r} \left( \because u = - 2r \right)$

$\Rightarrow v = \infty$
For the reflection in the concave mirror,
u = ∞
Thus, = focal length of mirror = r/2
For the refraction of APB of the reflected image,
u = −3r/2

$\Rightarrow \frac{1}{v} - \frac{1 . 5}{3r/2} = \frac{- 0 . 5}{- r} \left( Here, \mu_1 = 1 . 5 , \mu_2 = 1 and R = - r \right)$

$\Rightarrow v = - 2r$
As negative sign indicates images formed inside APB, so image should be at C.
Therefore, the final image is formed on the reflecting surface of the sphere.

Concept: Concave Mirror
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Chapter 18: Geometrical Optics - Exercise [Page 415]

APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 18 Geometrical Optics
Exercise | Q 46 | Page 415

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