IIT-JEE 2010 · Paper 1 · Physics Q75

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PhysicsInteger+3Question 75 of 84

The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25cm in front of it to 50cm, the magnification of its image changes from $m_{25}$ to $m_{50}$ . The ratio $\dfrac{m_{25}}{m_{50}}$ is

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Answer and worked solution

Official answer: 6

Magnification of a thin lens: $m=\dfrac f{f+u}$ (with $u$ negative):

$u=-25$ : $m_{25}=\dfrac{20}{20-25}=-4$ ;
$u=-50$ : $m_{50}=\dfrac{20}{20-50}=-\dfrac23$ .

Ratio. $\dfrac{m_{25}}{m_{50}}=\dfrac{-4}{-2/3}=6$ .

Answer: 6

Topics:Thin lenses and lens maker's formula

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IIT-JEE 2010 · Paper 1, Physics question 75