JEE PYQ: Ray Optics Question 21
Question 21 - 2020 (05 Sep Shift 1)
A compound microscope consists of an objective lens of focal length 1 cm and an eye piece of focal length 5 cm with a separation of 10 cm. The distance between an object and the objective lens, at which the strain on the eye is minimum is $\frac{n}{40}$ cm. The value of $n$ is _____.
Show Answer
Answer: (50)
Solution
Given: Length of compound microscope, $L = 10$ cm Focal length of objective $f_0 = 1$ cm and of eye-piece, $f_e = 5$ cm $u_0 = f_e = 5$ cm Final image formed at infinity ($\infty$), $v_e = \infty$ $v_0 = 10 - 5 = 5$ Using lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0} = \frac{1}{1}$ $\frac{1}{5} - \frac{1}{u_0} = \frac{1}{1} \Rightarrow u_0 = -\frac{5}{4}$ cm or, $\frac{5}{4} = \frac{N}{40}$ $\therefore N = \frac{200}{4} = 50$ cm.
BETA
