Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# A Paperweight in the Form of a Hemisphere of Radius 3.0 Cm is Used to Hold Down a Printed Page. an Observer Looks at the Page Vertically Through the - Physics

Sum

A paperweight in the form of a hemisphere of radius 3.0 cm is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the centre appear to the observer?

#### Solution

Given:
Radius of the paperweight (R) = 3 cm
Refractive index of the paperweight (μ2) = 3/2
Refractive index of the air (μ1) = 1

In the first case, the refraction is at A.
u = 0 and R = ∞
We know:

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

$\Rightarrow \frac{\frac{3}{2}}{v} - \frac{1}{0} = \frac{\mu_2 - \mu_1}{\infty}$

$\Rightarrow v = 0$
Therefore, the image of the letter is formed at the point.
For the second case, refraction is at point B.
Here,
Object distance, = −3 cm
= − 3 cm
μ1 = 3/2
μ2 = 1
Thus, we have:

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

$\frac{1}{v} - \frac{3}{2 \times ( - 3)} = \frac{1 - \frac{3}{2}}{- 3}$

$\Rightarrow \frac{1}{v} + \frac{3}{2 \times 3} = \frac{1}{6}$

$\Rightarrow \frac{1}{v} = \frac{1}{6} - \frac{1}{2}$

$\Rightarrow v = - 3 \text{ cm }$
Hence, there will be no shift in the final image.

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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 18 Geometrical Optics
Q 44 | Page 415