# NEET Physics Ray Optics and Optical Instruments Questions Solved

BOARD

Define magnifying power of a telescope. Write its expression.

A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece.                                                 5mark

a) Magnifying power is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.                                                                        1mark

Expression

m=$\beta }{\alpha }={f}_{0}}{{f}_{e}}$

1mark

[Award 1 mark if student writes expression with -ve sign]

b) Using, the lens equation for objective lens,:                                            1mark

$\frac{1}{{f}_{0}}=\frac{1}{{v}_{0}}=\frac{1}{{u}_{0}}$

1/2mark

$\approx 150cm$

Hence,  magnification due to the objective lens

${m}_{0}=\frac{{v}_{0}}{{u}_{0}}=\frac{150×{10}^{-2}m}{3000m}$                                                                  1/2mark

$\approx \frac{{10}^{-2}}{20}=.05×{10}^{-2}$

Using lens formula for eyepiece

$\frac{1}{{f}_{e}}=\frac{1}{{v}_{e}}-\frac{1}{{u}_{e}}$

$⇒\frac{1}{5}=\frac{1}{-25}-\frac{1}{{u}_{e}}$

$⇒\frac{1}{{u}_{e}}=\frac{1}{-25}-\frac{1}{5}=\frac{-1-5}{25}$

$⇒{u}_{e}=\frac{-25}{6}cm$

∴ Magnification due to eyepiece  ${m}_{e}=\frac{-25}{-\frac{25}{6}}=6$                                    1/2mark

Hence, total magnification => mme × mo

m= 6 X 5 × 10–4 = 30 × 10–4

Hence, size of final image

=  30 × 10–4 ×100 m

=  30 cm                                                                                            1/2mark

(Award full marks  for alternative method)

Difficulty Level:

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