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An object placed on a metre scale at $8 cm$ mark was focussed on a white screen placed at $92 cm$ mark using a converging lens placed on the scale at $50 cm$ mark. What is the focal length of the converging lens?

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+ 21 cm

- 40 cm

+ 35 cm

None of the above

answer is A.

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Detailed Solution

The Focal length of the lens,

f = + 21 cm

Given,
Positions of the object, lens and screen are at

8 cm

50 cm

, and

90 cm

marks respectively.
⇒

u = - (50 - 8)

= - 42 cm

⇒

v = (+ 92 - 50)

= + 42 cm

⇒

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

= \frac{1}{42} - \frac{1}{- 42}

= \frac{1}{42} + \frac{1}{42}

⇒

\frac{1}{f} = \frac{2}{42}

⇒

f = \frac{42}{2}

\Rightarrow f = + 21 cm

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