Question

A student focused the image of an object on a white screen using a converging lens. He noted down the positions of the object, screen and the lens on a scale as given below: Position of object= 10.0 cm; Position- of lens = 50.0 cm; Position of screen= 90.0 cm; (a) Find the focal length of the converging lens.; (b) Find the position of the image if the object is shifted towards the lens at a position of 30.0 cm.; (c) State the nature of the image formed if the object is further shifted towards the lens.(All India - 2013)

Answer 1

Answer:

the lens is covex lens...

Answer 2

Calculation related to a converging lens

Given:

Object distance (u) = -40 cm

Image distance (v) = 90 - 50 = 40 cm

To find:

Focal length

Position

Nature of the image

Formula to be used:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Calculation:

(a) Focal length of the converging lens

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f}=\frac{1}{40}-\frac{1}{-40}$

$\frac{1}{f}=\frac{2}{40}$

$\frac{1}{f}=\frac{1}{20}$

∴ The focal length of the converging lens is calculated as +20 cm.

(b) The position of the image if the object is shifted towards the lens at a position of 30.0 cm

Due to shifting, the new distance becomes -20 cm and focal length is +21 cm.

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

$\frac{1}{v}=\frac{1}{20}-\frac{1}{-20}$

$\frac{1}{v}=\frac{1}{0}$

∴ The value of v is zero, it means that the position of the object is at infinity.

(c) The nature of the image

• When the object is shifted from 30 cm to 50 cm, the new position of the object will be between the between focus and optical center of the lens.
• The nature of the formed image will be virtual, erect and enlarged in size and forms behind the lens.

Learn more about such concept

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