Most Repeated Questions in NEET Exam - Important Question for NEET 2024

Magnification In Lenses MCQ - Practice Questions with Answers

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

Quick Facts

  • Magnification in Lenses is considered one of the most asked concept.

  • 7 Questions around this concept.

Solve by difficulty

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in a normal setting, the angle formed by the image of the tower is \Theta, then \Thetaclose to :

A screen beaming a real image of magnification m1 formed by a convex lens is moved a distance \mathrm{x}. The object is moved until a new image of magnification \mathrm{m_{2}} is formed on the screen. The focal length of the lens is

Concepts Covered - 1

Magnification in Lenses

Magnification in Lenses-

Magnification produced by a lens is defined as the ratio of size of the image to that of the object. 

                                                   

For the derivation let an object is palced whose height is h and the image height is h'. Let us consider an object is placed on the principal axis with its height h perpendicular to the principal axis as shown in figure. The first ray is passing through the optic centre will go undeviated. The second ray parallel to the principal axis must pass through the focus F' The image is formed where both the light rays intersect. By this we get an image of height h'. So mathematically, magnification can be written as - 

                                                                       

$$
\mathbf{m}=\frac{\mathbf{h}^{\prime}}{\mathbf{h}}
$$


Like mirror we can derive the formula of magnification in terms of the position of object and image. For this, consider the figure given below -

As the triangle OO'P and II'P, is similar so we can write that -

$$
\frac{I^{\prime} I}{O O^{\prime}}=\frac{P I}{P O}
$$


So,

$$
\frac{h^{\prime}}{h}=\frac{v}{u}
$$


$$
m=\frac{h^{\prime}}{h}=\frac{v}{u}
$$
 

Study it with Videos

Magnification in Lenses

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top