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Magnification In Lenses - Practice Questions & MCQ

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.

  • 9 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 :

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are:

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

An object is placed at a distance of 10 \mathrm{~cm} to the left on the axis of a convex lens A of focal length 20 \mathrm{~cm}. A second convex lens of focal length 10 \mathrm{~cm} is placed co-axially to the right of the lens A at a distance of 5 \mathrm{~cm} from A. Find the position of the final image and its magnification. Trace the path of the rays.   

Find the size of the image formed in the situation shown in figure.

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{h'}{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'I}{OO'} = \frac{PI}{PO}

                                                                       So, 

                                                                                \frac{h'}{h} = \frac{v}{u}

                                                                          

                                                                             m = \frac{h'}{h} = \frac{v}{u}

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Magnification in Lenses

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