A glass slab of thickness 20, cm and refracti | vedclass

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Question

(A) When light from an object passes through a glass slab,it undergoes a normal shift given by $x = d \left( 1 - \frac{1}{\mu} \right)$.Given $d = 20\

Vedclass · Accepted Answer

$\frac{100}{3}\, cm$

Vedclass · Answer

(A) When light from an object passes through a glass slab,it undergoes a normal shift given by $x = d \left( 1 - \frac{1}{\mu} \right)$.Given $d = 20\, cm$ and $\mu = 1.5 = \frac{3}{2}$,the shift is:$x = 20 \left( 1 - \frac{1}{1.5} \right) = 20 \left( 1 - \frac{2}{3} \right) = 20 \left( \frac{1}{3} \right) = \frac{20}{3}\, cm$.This shift is towards the mirror,so the effective position of the object relative to the mirror becomes $u = 40 - x = 40 - \frac{20}{3} = \frac{120 - 20}{3} = \frac{100}{3}\, cm$.The plane mirror forms an image at the same distance behind it as the object is in front of it.Therefore,the image is formed at a distance of $\frac{100}{3}\, cm$ behind the mirror.

$A$ glass slab of thickness $20\, cm$ and refractive index $1.5$ is placed in front of a plane mirror. An object is placed at a distance of $40\, cm$ from the mirror in air. The position of the final image with respect to the mirror will be at a distance of:

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