Question

A microscope is having objective of focal length $1\,cm$ and eyepiece of focal length $6 \,cm$ If tube length is $30\,cm$ and image is formed at the least distance of distinct vision, what is the magnification produced by the microscope ? Take $D \,= \,25\, cm$

• A. 6
• B. 150
• C. 25
• D. 125

Answer 1

The Correct Option is B

By compound microscope
$m=\frac{L}{f_{0}}\left(1+\frac{D}{f_{e}}\right)$
(where length of tube $L=30 cm$, focal length of objective lens $f_{0}=1 cm$, focal length of
eye-piece $f_{e} =6\, cm,\, D=25\, cm)$
$=\frac{30}{1}\left(1+\frac{25}{6}\right)=30 \times \frac{(6+25)}{6}$
$=5 \times 31$
$=155 cm \cong 150$

Answer 2

The microscope is an optical instrument which is used for magnifying very tiny objects that are not possible to see by the naked eye such as biological specimens and samples. The microscope, as the name suggests, is used to view the objects whose size is in the range of microns (10^-6m).

There are various types of microscopes such as Optical microscope, fluorescence microscope, X-ray microscope and scanning electron microscope.

Firstly, let us understand the working of an optical microscope.

The optical microscope contains 2 lenses, such as: Objective and Eyepiece.

 The object that has to be viewed is placed on the platform with light shining beneath it, as shown in the above figure. The objective lens is a biconvex lens that is responsible to produce a real image inside the tube of the microscope. Thus, the image formed becomes the object for another lens called eyepiece through which the human eye perceives the image. The image formed by this eyepiece is virtual, erect and formed beyond the object at a distance known as distance of least vision, up to where the normal human eye can actually perceive the complete image.

Hence, the magnification of the objective lens is the ratio of the length of the microscope tube at the end of which the eyepiece is present, to its focal length.

Magnification, \(m_{1}=-\frac{L}{f_{1}}\)

So, here the given values are,

Length of the tube, \(L\)=30cm 

Focal length, \(f_{1}\)=1cm

After substituting the values,

\(m_{1}=-\frac{L}{f_{1}}\) = \(\frac{30}{1}\) 

Likewise, the magnification of the eyepiece is the ratio of the largest distance that the human eye can perceive, to its focal length is:

Magnification, \(m_{2}=-\frac{D}{f_{2}}\)

So, here the given values are,

Length of the tube, \(D\)=25cm

Focal length, \(f_{2}\)=6cm

After substituting the values,

\(m_{2}=-\frac{D}{f_{2}}\) = \(\frac{25}{6}\) 

 Hence, the total magnification is equal to the product of magnification of both the lenses.

Therefore, total magnification, m = m₁m₂ = \(\frac{30}{1}\times(1+\frac{25}{6})\)

= \(30\times\frac{(6+25)}{6}\)

= \(5\times31\)

= \(155\) ≅ \(150\)

Hence, the magnification is equal to, 

\(|m|=150\)