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Question: A square card of side length 1 mm is being seen through a magnifying lens of focal length of focal l...

A square card of side length 1 mm is being seen through a magnifying lens of focal length of focal length 10 cm. The card is placed at a distance of 9cm from the lens. The card is placed at a distance of 9 cm from the lens. The apparent area of the card through the lens is :

A

1 cm21 \mathrm {~cm} ^ { 2 }

B

0.81 cm20.81 \mathrm {~cm} ^ { 2 }

C

0.27 cm20.27 \mathrm {~cm} ^ { 2 }

D

0.60 cm20.60 \mathrm {~cm} ^ { 2 }

Answer

1 cm21 \mathrm {~cm} ^ { 2 }

Explanation

Solution

: Area of a square card = 1 mm ×1 mm = 1 mm2\mathrm { mm } ^ { 2 }

Focal length of magnifying lens (converging lens),

F = + 10 cm

Object distance, u = - 9 cm

According to thin lens formula

1v=1f+1u=1+10 cm+1−9 cm=110 cm−19 cm\frac { 1 } { v } = \frac { 1 } { \mathrm { f } } + \frac { 1 } { \mathrm { u } } = \frac { 1 } { + 10 \mathrm {~cm} } + \frac { 1 } { - 9 \mathrm {~cm} } = \frac { 1 } { 10 \mathrm {~cm} } - \frac { 1 } { 9 \mathrm {~cm} }

or v=−90 cm\mathrm { v } = - 90 \mathrm {~cm}

magnification,m=vu=−90 cm−9 cm=10\mathrm { m } = \frac { \mathrm { v } } { \mathrm { u } } = \frac { - 90 \mathrm {~cm} } { - 9 \mathrm {~cm} } = 10

∴\therefore Apparent area of the card through the lens

=10×10×1 mm2= 10 \times 10 \times 1 \mathrm {~mm} ^ { 2 }

=100 mm2=1 cm2= 100 \mathrm {~mm} ^ { 2 } = 1 \mathrm {~cm} ^ { 2 }