Question

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 ?

Answer 1

Given :

Refraction at A( first surface):

μ2=3/2

μ1= 1

u= 0

R= ∞

From the formula

μ2/V- μ1/u= μ2-μ1/R

V=0

since u=o and R=∞

Second refraction at B:

u=-3cm

R=-3cm

μ1=3/2

μ2=1

1/v+3/2x3=1-1.5/-3 =1/6

1/v=1/6-1/2=-1/3

V=-3cm

There will  not be any shift in the final image

Answer 2

Height above the page will the printed letters near the centre appear to the observer is - 3 cm.

Explanation:

Given data in the question  :

Step 1 :

A(first surface refraction)

$\mu_{2}=\frac{3}{2}$

$\mu_{1}=1$

U = 0

R = ∞

We know that  

$\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$

V = 0

Since, u = o and R = ∞

Step 2 :

Now for the second refraction at B, the image will be created at the point

u = -3 cm

R = -3 cm

$\mu_{1}=\frac{3}{2}$

$\mu_{2}=1$

$\frac{1}{v}+\frac{3}{2 \times 3}=\frac{1-1.5}{-3}$

$\frac{1}{v}+\frac{3}{6}=\frac{-0.5}{-3}$

$\frac{1}{v}+\frac{1}{2}=\frac{5}{30}$

$\frac{1}{v}+\frac{1}{2}=\frac{1}{6}$

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

$\frac{1}{v}=\frac{1-3}{6}$

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

v = -3

Therefore, there will not be any shift in the final image .