MP NEET PG 2025 Round 1 Allotment: Counselling Revised Merit List

Simple Microscope MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • 3 Questions around this concept.

Solve by difficulty

The eye-piece and objective of a microscope, having focal lengths of 0.3 m and 0.4 m respectively, are separated by a distance of 0.2 m. Now the eyepiece and the objective are to be interchanged such that the angular magnification of the instrument remains the same. The new separation between the lenses is:

The focal length of the object glass of a microscope is 2 cm, that of the eye-piece is 4 cm and the distance between them is 20 cm. What should be the object distance from the object-glass if the final image is to be seen at least a distance of distinct vision (25 cm) from the eyepiece? What will be the magnification in this situation?

Concepts Covered - 1

Simple Microscope

Magnifying power: It is defined as the ratio between the dimensions of the image and the object.  

Magnifying power of an optical instrument is given by, 

$m=\frac{\text { Visual angle with instrument }(\beta)}{\text { Visual angle when object is placed at least distance of distinct vision }(\alpha)}$

                                                                          

 

Simple Microscope

  • It is a single convex lens of lesser focal length.
  • Also called magnifying glass or reading lens.

 

Case 1: Magnification's, when the final image is formed at D and $\infty$.

$$
\begin{aligned}
& \text { (i.e. } m_D \text { and } m_{\infty} \text { ) } \\
& m_D=\left(1+\frac{D}{f}\right)_{\max \text { and }} \\
& m_{\infty}=\left(\frac{D}{f}\right)_{\min }
\end{aligned}
$$


Case 2: If lens is kept at a distance a from the eye then

$$
\begin{aligned}
& m_D=1+\frac{D-a}{f} \text { and } \\
& m_{\infty}=\frac{D-a}{f}
\end{aligned}
$$

 

 

 

 

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Simple Microscope

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