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Question: A beam of light is converging towards a point \(t\), refractive index = \(\mu\), is introduced in th...

A beam of light is converging towards a point tt, refractive index = μ\mu, is introduced in the path of the beam. The convergence point is shifted by

A

t(11μ)t \left( 1 - \frac { 1 } { \mu } \right) away

B

t(1+1μ)t \left( 1 + \frac { 1 } { \mu } \right) away

C

t(11μ)t \left( 1 - \frac { 1 } { \mu } \right) nearer

D

t(1+1μ)t \left( 1 + \frac { 1 } { \mu } \right) nearer

Answer

t(11μ)t \left( 1 - \frac { 1 } { \mu } \right) away

Explanation

Solution

Normal shift Δx=(11μ)t\Delta x = \left( 1 - \frac { 1 } { \mu } \right) t

and shift takes place in direction of ray.