Question

A compound microscope consists of an eyepiece lens of focal length 5 cm and objective lens of focal length 2 cm separated by 15 cm. If d cm is the distance of object from objective lens so that the image is obtained at infinity. Find value of 2d.

Answer 1

40/9

Answer 2

For the final image to be formed at infinity, the object for the eyepiece must be placed at its focal point. This object is the intermediate image formed by the objective lens.

Let $f_o$ be the focal length of the objective lens, $f_e$ be the focal length of the eyepiece lens, and $L$ be the distance between the lenses. Given: $f_o = 2$ cm, $f_e = 5$ cm, $L = 15$ cm.

For the final image to be at infinity, the object distance for the eyepiece ($u_e$) must be $-f_e$. So, $u_e = -5$ cm.

The intermediate image formed by the objective lens is at a distance $v_o$ from the objective. This image acts as the object for the eyepiece. The distance relationship is $u_e = L - v_o$. Substituting the values: $-5 \text{ cm} = 15 \text{ cm} - v_o$. This gives the image distance from the objective lens: $v_o = 15 \text{ cm} + 5 \text{ cm} = 20$ cm.

Now, using the lens formula for the objective lens, $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$, where $u_o$ is the object distance from the objective (given as $d$). We take $u_o = -d$ as the object is placed to the left.

$\frac{1}{20 \text{ cm}} - \frac{1}{-d} = \frac{1}{2 \text{ cm}}$ $\frac{1}{20} + \frac{1}{d} = \frac{1}{2}$ $\frac{1}{d} = \frac{1}{2} - \frac{1}{20} = \frac{10 - 1}{20} = \frac{9}{20}$ $d = \frac{20}{9}$ cm.

The question asks for the value of $2*d$. $2*d = 2 \times \frac{20}{9} = \frac{40}{9}$ cm.