Question

Two beams of light are incident normally on water (R.I. = 4/3). If the beam 1 passes through a glass of height h as shown in the figure, the time difference for both the beams for reaching the bottom is

• A. Zero
• B.
• C.
• D. $\frac { h } { 6 C }$

Answer 1

Answer: (D)

$\frac { h } { 6 C }$

Answer 2

The refractive index of glass is greater than that of water. Therefore the speed of light in glass is lesser than that of water. It is given as

v = $\frac { C } { n }$ where C = 3 x 10⁸

v = speed of light in a medium of R.I. n

∴The time difference for the rays
= t₁ – t₂ = $\frac { h } { v _ { g } } - \frac { h } { v _ { w } }$

⇒ ∆t =

⇒ ∆t = $\frac { \mathrm { h } } { \mathrm { C } } \left( \frac { 3 } { 2 } - \frac { 4 } { 3 } \right) = \frac { \mathrm { h } } { 6 \mathrm { C } }$