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=βα=f0fe

or m=f0fe1+feD                                                                                  1mark

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

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

1f0=1v0=1u0                      

      1150=1v0-1-3×105

       1v0=1150-1-3×105=2000-13×105                                                1/2mark

   v0=-3×1051999cm

150cm

Hence,  magnification due to the objective lens

     m0=v0u0=150×10-2m3000m                                                                  1/2mark

10-220=.05×10-2

Using lens formula for eyepiece

1fe=1ve-1ue

15=1-25-1ue

1ue=1-25-15=-1-525

ue=-256cm

∴ Magnification due to eyepiece  me=-25-256=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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