Question

A microscope is having an objective of focal length 1 cm and eyepiece of focal length 6 cm. If tube length is 30cm and image is formed at the least distance of distant vision, what is the magnification produced by the microscope? Take D = 25cm
A. 6
B. 150
C. 25
D. 125

Answer 1

The microscope has two lenses: one is near the object called objective and the other is near the eye called eyepiece. The total magnification produced by the microscope is equal to the product of individual magnifications of both the objective and eyepiece lens.

Complete step-by-step answer:
The microscope is an optical instrument used for magnifying very tiny objects that are not seen 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 $\left( {{{10}^{ - 6}}m} \right)$ .

There are several types of microscopes such as Optical microscope, fluorescence microscope, X-ray microscope and scanning electron microscope.
Here, let us understand the working of an optical microscope.
The optical microscope contains 2 lenses – Objective and Eyepiece.
The object 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 produces a real image inside the tube of the microscope. 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.
Thus, 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, ${m_1} = - \dfrac{L}{{{f_1}}}$
Given,
Length of the tube, $L = 30cm$
Focal length, ${f_1} = 1cm$
Substituting,
${m_1} = - \dfrac{L}{{{f_1}}} = - \dfrac{{30}}{1} = - 30$
Similarly, the magnification of the eyepiece is the ratio of the largest distance that the human can perceive, to its focal length.
Magnification, ${m_2} = \dfrac{D}{{{f_2}}}$
Given,
Length of the tube, $D = 25cm$
Focal length, ${f_2} = 6cm$
Substituting,
${m_2} = \dfrac{D}{{{f_2}}} = \dfrac{{25}}{6}$
The total magnification is equal to the product of magnification of both the lenses.
Total magnification, $m = {m_1}{m_2} = - 30 \times \dfrac{{25}}{6} = - 5 \times 25 = - 125$
Hence, the magnification is equal to, $\left| m \right| = 125$

Hence, the correct option is Option D.

Note: The students can see that in the case of the objective lens, the magnification has negative sign and, in the eyepiece, the magnification has positive sign.
This is because the negative sign of the magnification indicates that the image is real and the positive sign of it, indicates the virtual image.
In the answer, we have got the magnification as – 125. Hence, the image produced by the microscope is real.