The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ______ cm.

Question

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ______ cm. [2026]

Smart Achievers · Accepted Answer

(16)

$m = \frac{f}{f + u}$

$m_{1} = - m_{2}$

$\frac{f}{f - 8} = - \frac{f}{f - 24}$

$f - 8 = 24 - f$

$2 f = 32$

$f = 16 cm$

Solution

Q.
The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ______ cm. [2026]

Ans.
(16)

$m = \frac{f}{f + u}$

$m_{1} = - m_{2}$

$\frac{f}{f - 8} = - \frac{f}{f - 24}$

$f - 8 = 24 - f$

$2 f = 32$

$f = 16 cm$

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Solution

Q. The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ______ cm. [2026]

Ans. (16) m=ff+u m1=-m2 ff-8=-ff-24 f-8=24-f 2f=32 f=16 cm

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Q.
The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ______ cm. [2026]

Ans.
(16)

$m = \frac{f}{f + u}$

$m_{1} = - m_{2}$

$\frac{f}{f - 8} = - \frac{f}{f - 24}$

$f - 8 = 24 - f$

$2 f = 32$

$f = 16 cm$