Question 285 of 295

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

A$30^{\circ}$B$15^{\circ}$C$1^{\circ}$D$86^{\circ}$

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Answer: (D) $86^{\circ}$

Angular magnification of a telescope in normal adjustment is
$m=\dfrac{\beta }{\alpha }=\dfrac{{f}_{0}}{{f}_{e}}\Rightarrow \beta =\dfrac{{f}_{0}}{{f}_{e}}\alpha$



According to the question
$\alpha =\dfrac{y}{x}=\dfrac{50}{1000}=\dfrac{1}{20}$
$\Rightarrow \alpha =\dfrac{1}{20}$
$\Rightarrow \beta =\dfrac{150}{5}\times \dfrac{1}{20}=32rad$
$=\dfrac{3}{2}\times \dfrac{180^{\circ}}{3.14}=86^{\circ}$