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 \Theta, then \Theta is close to :

  • Option 1)

    1^{\circ}

  • Option 2)

    15^{\circ}

  • Option 3)

    30^{\circ}

  • Option 4)

    60^{\circ}

 

Answers (1)

As we learnt in

Magnification from mirror -

m= \frac{h_{i}}{h_{o}}= \frac{-v}{u}
 

- wherein

h_{i}= height of image from principal axis

h_{o}= height of object from principal axis

u= object distance

v= image distance

 

 Angular magnification m=\frac{f_{o}}{f_{e}}=\frac{150}{5}=30

\therefore\ \; \frac{tan\beta}{tan\alpha}=30\ \; \Rightarrow\ \; tan\beta=tan\alpha \times 30

tan\beta= \left(\frac{50}{1000}\times30 \right )=\frac{15}{10}=\frac{3}{2}

tan\beta=\frac{3}{2}\ \; \Rightarrow\ \;\tan\beta=tan^{-1}\left(\frac{3}{2} \right )

\theta=\beta \simeq60^{o}

Correct option is 4.

 


Option 1)

1^{\circ}

This is an incorrect option.

Option 2)

15^{\circ}

This is an incorrect option.

Option 3)

30^{\circ}

This is an incorrect option.

Option 4)

60^{\circ}

This is the correct option.

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