CBSE Class 12 · 2005 · Geometrical OpticsShort Answer

A figure divided into squares, each of size

1 mm^{2}

, is being viewed at a distance of

9 cm

through a magnifying lens of focal length

10 cm

, held close to the eye.
(a) Draw a ray diagram showing the formation of the image.
(b) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(c) What is the angular magnification of the lens?

Solution

With

u = - 9 cm

f = + 10 cm

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{9} = - \frac{1}{90} \Rightarrow v = - 90 cm

(b)

m = \frac{v}{u} = \frac{- 90}{- 9} = 10

. Each square of area

1 mm^{2}

appears as

(10)^{2} \times 1 = 100 mm^{2}

.
(c) Angular magnification

= \frac{D}{u} = \frac{25}{9} \approx 2.8

(object nearer than the near point, so the lens lets the eye view it comfortably with magnification).

CBSE Class 12 Board · 2005 · Subjective

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CBSE Class 12 · 2005 · Geometrical OpticsShort Answer

Solution

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