A simple microscope of focal length \(f\) forms the final image at infinity. To find its angular magnification, an object of height \(h\) is placed at the focal point of the lens. The angle subtended at the eye by the image is about
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\frac{h}{f}\)
The eye is kept close to the lens. So the image subtends the same angle as the object does at the lens. For a small angle, this angle is the height divided by the distance.
\( u=-f \)
\( \begin{aligned} \theta_i &\approx \frac{h}{-u} \\ &= \frac{h}{f} \end{aligned} \)
Without the lens, the object is seen best at the near point, where it subtends \( \theta_o\approx h/D \). The ratio of the two angles gives the angular magnification.
\( \begin{aligned} m &= \frac{\theta_i}{\theta_o} \\ &= \frac{D}{f} \end{aligned} \)
Hence, the image subtends an angle of about \( h/f \) at the eye.
NCERT (2026-27):
The microscope – “We now find the angle subtended at the eye by the image when the object is at u. … we have the angle subtended by the image”
\( \displaystyle \tan\theta_i=\frac{h^{\prime}}{-v}=\frac{h}{-v}\cdot\frac{v}{u}=\frac{h}{-u}\approx\theta \)
“The angle subtended by the object, when it is at u = –f.”
\( \displaystyle \theta_i=\left(\frac{h}{f}\right) \)\( \qquad (9.41) \)
“… The angular magnification is, therefore”
\( \displaystyle m=\left(\frac{\theta_i}{\theta_o}\right)=\frac{D}{f} \qquad (9.42) \)