A beam of light is converging towards a point | vedclass

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(A) When a parallel-sided glass slab of thickness $t$ and refractive index $\mu$ is placed in the path of a converging beam of light,the light rays un

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

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(A) When a parallel-sided glass slab of thickness $t$ and refractive index $\mu$ is placed in the path of a converging beam of light,the light rays undergo refraction at both surfaces of the slab.Due to refraction,the rays bend towards the normal when entering the glass and away from the normal when exiting.This causes the point of convergence to shift in the direction of the incident light rays.The magnitude of this normal shift $\Delta x$ is given by the formula:$\Delta x = t \left( 1 - \frac{1}{\mu} \right)$Since the shift occurs in the direction of the light beam,the new convergence point $I'$ moves away from the slab compared to the original point $I$.

$A$ beam of light is converging towards a point $I$ on a screen. $A$ plane glass plate of thickness $t$ and refractive index $\mu$ is introduced in the path of the beam. The convergence point is shifted by:

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