Question

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Answer 1

QUESTION:

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ , then θ is close to

ANSWER:

• θ is close to 60°.

GIVEN:

• Focal length of telescope(F) = 150 cm

• Focal length of eyepiece(f) = 50 cm

• Distance from the tower(d) = 1000m

• Height of the tower = 50m

TO FIND:

• CLOSEST VALUE OF θ

FORMULA:

Magnification from mirror:

$Magnifying\:\:distance(m)= F/f$

EXPLANATION:

d = 1000 m

F = 150 cm

f = 50 cm

$Magnifying\:\:distance =150 / 5= 30\\ \\ \\ Magnifying\:\:distance = \dfrac{\tan\beta}{\tan\alpha } = 30\\ \\ \\\tan \alpha = \dfrac{f}{d}\\ \\ \\ \tan \alpha = \dfrac{50}{1000} \\ \\ \\ \tan \alpha = \dfrac{1}{20} \\ \\ \\ \tan \beta = m\times \tan \alpha \\ \\ \\ \tan \beta = 30\times \dfrac{1}{20} \\ \\ \\ \tan\beta = \dfrac{30}{20} \\ \\ \\ \tan \beta = \dfrac{3}{2}\:\:radians \\ \\ \\ \beta =tan^{-1}\bigg( \dfrac{3}{2}\bigg) \\ \\ \\ \beta \approx 60^{ \circ}$

$\bold{\large{NOTE 1 : \tan \alpha= Angle\:\:formed\:\:by\:\:the \:\:eyepiece}}$

$\bold{\large{NOTE 2 :\tan \beta =Angle\:\:formed\:\:by\:\:the \:\:image}}$

OTHER FORMULA:

$m= H/h= -v/ u\\ \\ \\H= Height\:\:of\:\: image\\ \\\\h= Height\:\:of\:\:object\\ \\ \\u= Object\:\:distance\\ \\ \\v= Image\:\:distance$

Answer 2

We know,

Resolving Power = θ=

D

d

but θ≈

a

λ

(Note:exact relation,θ=

a

1.22

λ)

D

d

=

a

λ

⇒=d=

a

λD

⇒d=

10×10

−2

5000×10

−10

×10

3

=5mm