Ray Optics

JEE Main level, single correct, medium

200 of 295

Question 200 of 295

A convex lens is put 10cm10cm from a light source and it makes a sharp image on a screen, kept 10cm10cm from the lens. Now a glass block (refractive index 1.5) of 1.5cm1.5cm thickness is placed in between the light source and the lens. To get the sharp image again, the screen is shifted by a distance d.d. Then dd is:

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Answer: (B) 0.55cm0.55cm away from the lens

From given data

1v−1u=1f\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

110−1(−10)=1f\dfrac{1}{10}-\dfrac{1}{(-10)}=\dfrac{1}{f}

f=5cmf=5cm

When glass plate is placed, shift in object for lens

s=t(1−1μ)=1.5(1−23)s=t\left(1-\dfrac{1}{\mu }\right)=1.5\left(1-\dfrac{2}{3}\right)

=0.5cm=0.5cm

Shift will be in the direction of incoming rays

∴u=10−0.5=9.5cm∴u=10-0.5=9.5cm

1v−1u=1f\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

1v=15+1(−9.5)\dfrac{1}{v}=\dfrac{1}{5}+\dfrac{1}{(-9.5)}

v=47.54.5cmv=\dfrac{47.5}{4.5}cm

Shift =47.54.5−10=2.54.5=59cm=0.55cm=\dfrac{47.5}{4.5}-10=\dfrac{2.5}{4.5}=\dfrac{5}{9}cm=0.55cm away from lens.

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