Question

in compound microscope the length of microscope is 21.5 CM focal length of objective is 1.6 CM focal length of eyepiece is 2.1 cm and A final image is formed at Infinity that distance of object from objective lens is​

Answer 1

The distance of object from objective lens is​ 1.744 cm

Explanation:

Given:

Focal length of objective $f_o=1.6$ cm

Focal length of eyepiece $f_e=2.1$ cm

Length of the microscope $L=21.5$ cm

To find out:

Distance of the object fro objective lens

i.e. $u_o=?$

Solution:

The image formed by the objective lens will act as object for eyepiece and its image is formed at infinity

Therefore,

$L=v_o+u_e$

Since image from eyepiece is at infinity

Therefore, the image formed by the objective lens must be at the focus of the eyepiece

Thus,

$L=v_o+f_e$

$\implies 21.5=v_o+2.1$

$\implies v_o=21.5-2.1$

$\implies v_o=19.4$ cm

Using lens formula

$\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$

$\frac{1}{19.4}-\frac{1}{u_o}=\frac{1}{1.6}$

$\implies -\frac{1}{u_o}=\frac{1}{1.6}-\frac{1}{19.4}$

$\implies -\frac{1}{u_o}=\frac{1}{4}(\frac{1}{0.4}-\frac{1}{4.85})$

$\implies -\frac{1}{u_o}=\frac{1}{4}(\frac{4.85-0.4}{0.4\times 4.85})$

$\implies -\frac{1}{u_o}=\frac{1}{4}(\frac{4.45}{1.94})$

$\implies -u_o=\frac{4\times 1.94}{4.45}$

$\implies -u_o=\frac{4\times 1.94}{4.45}$

$\implies u_o=-1.744$ cm

Negative sign indicates that the object is on the opposite side of the image formed by the lens

Hope this answer is helpful.

Know More:

Q: A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25 cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Click Here: https://brainly.in/question/6095664