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A 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 index 1.5) 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 :

(1) 0.55 cm away from the lens

(2) 1.1 cm away from the lens

(3) 0.55 cm towards the lens

(4) 0
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Ans: (1) 0.55 cm away from the lens

Sol: 

${1\over v}-{1\over u}={1\over f}\implies{1\over10}-{1\over-10}={1\over f}$

$\implies f=5cm$

Shift due to slab $=t\left(1-{1\over\mu}\right)$ in the direction of incident ray

$=1.5\left(1-{2\over3}\right)=0.5$

again, ${1\over v}-{1\over-9.5}={1\over5}$

$\implies{1\over v}={1\over5}-{2\over19}={9\over95}$

$\implies v={95\over9}=10.55cm$

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