Question

How can the resolving power of a telescope be increased? A telescope has an objective of
focal length 50 cm and an eyepiece of focal length 5 cm. The least distance of distinct
vision is 25 cms. The telescope is focused for distinct vision on a scale 200 cms away from
the objective. Calculate (i) the separation between objective and eyepiece (ii) the
magnification produced.​

Answer 1

Answer :

Given u=−200cm,f=50cm

For image I

1

of object formed by objective lens,

f

1

=

v

1

−

u

1

We have

v

1

=

f

1

+

u

1

=

50

1

+

200

1

=

200

4−1

=

200

3

⇒v=+

3

200

cm

Also, magnification produced by objective lens

m

0

=

u

v

=−

200

200/3

=

3

1

Image I

1

acts as an object for eye lens.

Here, v=−25cm,f=5cm

∴

f

1

=

v

1

−

u

1

⇒

u

1

=

v

1

−=

f

1

=−

25

1

−

5

1

=−

25

1+5

∴u=−

6

25

cm

And magnification produced by eye lens,

m

e

=

u

v

=

(−25/6)

−25

=6

a. The separation between objective and eyepiece

=∣V∣+∣u∣=

3

200

+

6

25

=

6

425

=70.73cm

b. Magnification produced, m=m

0

×m

e

=−

3

1

×6=−2

The negative sign shown that the final image is inverted.

Sorry for the amount of spaces

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