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Question

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 on its other side. 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 must be shifted by a distance d. Find the value of d.

A
1.1 cm away from the lens
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B
0
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C
0.55 cm towards the lens
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D
0.55 cm away from the lens
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Solution

The correct option is D 0.55 cm away from the lens
Given, 2f=10 cmf=5 cm

Now, due to glass slab,

Shift (S)=t(11μ)

Substituting the given data we get,

S=1.5(123)=0.5 cm

According to the sign convention,

u=(100.5)=9.5 cm

Now, From lens formula

1v1u=1f

Substituting the values of v,u and f

v=(5)(9.5)4.5=47.54.5=10.55 cm

Hence the screen should be shifted by 0.55 cm.



Hence, option (d) is the correct answer.

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