CBSE Class 12 · 2003 · Geometrical OpticsShort Answer

A converging lens of focal length

6.25 cm

is used as a magnifying glass. If the near point of the observer is

25 cm

from the eye and the lens is held close to the eye, calculate
(i) the distance of the object from the lens,
(ii) the angular magnification (magnifying power). Also find the angular magnification when the final image is formed at infinity.

Answer: (i) $u = - 5 cm$ ;
(ii) $M = 5$ (image at near point); $M = 4$ when the final image is at infinity.

Solution

(i) Image at the near point,

v = - 25 cm

f = + 6.25 cm

\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{- 25} - \frac{1}{u} = \frac{1}{6.25} \Rightarrow u = - 5 cm

(ii) Image at near point:

M = 1 + \frac{D}{f} = 1 + \frac{25}{6.25} = 5

.
Image at infinity:

M = \frac{D}{f} = \frac{25}{6.25} = 4

CBSE Class 12 Board · 2003 · Subjective

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