Question

The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in cm of the object and the image produced by the objective measured from the objective lens are respectively
A.$2.4$and$12.0$
B. $2.4$and$15.0$
C. $2.3$and$12.0$
D. $2.3$and$3.0$

Answer 1

Hint:
The distance between the objective lens and the eyepiece lens is measured as the length of the microscope tube. It is based on the object's position and the eyepiece lens's focal length. You can utilize the lens manufacturer's formula, which links image distance, object distance, and focal length.


Formula used:
$\dfrac{1}{{{v_0}}} - \dfrac{1}{{{u_0}}} = \dfrac{1}{{{f_0}}}$

Complete step by step solution:
Using a microscope, one can view things or creatures that are invisible to the naked eye. A microscope is used to magnify incredibly small objects. The objective lens and the eyepiece are the two convex lenses found on a compound microscope. Both impact the overall magnification.
Given the focal length of the objective lens is ${f_0} = 2.0cm$ , the focal length the eye-piece of a compound microscope ${f_e} = 3cm$
Since the final image formed by the eye-piece is at infinity.
Length of tube is given by distance between the objective and the eye-piece is $l = 15cm$
[It is given that the Image formed by the eye piece is at infinity, it means the object for the eye piece is at focus ${f_e}$ of the eye piece. This object will act as an image for objective ${v_0}$ ]

$ \Rightarrow l = {v_0} + {f_e}$
$ \Rightarrow 15 = {v_0} + 3$
$ \Rightarrow {v_0} = 12cm$

Now we can use the lens makers formula for the objective lens.
$\dfrac{1}{{{v_0}}} - \dfrac{1}{{{u_0}}} = \dfrac{1}{{{f_0}}}$
$ \Rightarrow \dfrac{1}{{12}} - \dfrac{1}{{{u_0}}} = \dfrac{1}{2}$
$ \Rightarrow {u_0} = - 2.4cm$
Hence the answer is option A.



Note:
While solving these types of questions, take care of sign convention used in lens maker formula. The object produces an image with certain magnification by the objective lens with ${f_0}$ , this image now acts as a virtual object to the eyepiece with ${f_e}$ which in turn produces a virtual image, which is magnified further.