JEE Advance - Physics (2010 - Paper 2 Offline - No. 11)
A large glass slab ($$\mu$$ = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?
Answer
6
Explanation
From the figure shown here, we have
$$\tan {i_c} = {R \over t}$$
$$\sin {i_c} = {R \over {\sqrt {{R^2} + {t^2}} }} = {1 \over \mu } = {3 \over 5}$$
$$25{R^2} = 9{R^2} + 9{t^2}$$
$$16{R^2} = 9{t^2} \Rightarrow R = {{3t} \over 4} = {{3 \times 8} \over 4} = 6$$ cm
Comments (0)
