# Convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is:-

Refraction through parallel slab -

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

- wherein

$s=$  shifting of object from slab

$t=$ thickness of slab

$\mu =$ Refractive Index of slab.

F = $\frac{10}{2}$ = 5 cm

Due to glass plate shift

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

= 1.5 $\left ( 1 - \frac{2}{ 3 } \right )$

= 0.5

New u =9.5 cm

Using lens formula

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

$\frac{1}{5} + \frac{1}{-9.5}$

$\frac{1}{v} = \frac{+4.5}{+9.5\times 5}$

v = 10.55 cm

Shift  is 0.55 cm away from the lens.

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