Question

. An object is placed at a distance of 2.5 cm form a screen and a convex lens is interposed between them. The magnification produced is 4. What is the focal length of the lens?

Please answer correctly ​

Answer 1

SOLUTION

GIVEN

• An object is placed at a distance of 2.5 cm form a screen and a convex lens is interposed between them.

• The magnification produced is 4.

TO DETERMINE

The focal length of the lens

CONCEPT TO BE IMPLEMENTED

The linear magnification is related to Image distance and Object distance as below

$\displaystyle \sf{Linear \: magnification = m = \frac{Image \: distance}{Object \: distance} = \frac{v}{u} }$

EVALUATION

Let image distance = v

Object distance = u

Now it is given that the magnification produced is 4.

$\displaystyle \sf{ \frac{v}{u} = - 4 }$

$\displaystyle \sf{ \implies \: u = - \frac{v}{4} \: \: \: - - - - (1)}$

Now it is given that object is placed at a distance of 2.5 cm form a screen and a convex lens is interposed between them.

$\sf{ |u| + |v| = 2.5 }$

$\displaystyle \sf{ \implies \: \frac{v}{4} + v= 2.5 }$

$\displaystyle \sf{ \implies \: \frac{5v}{4} = 2.5 }$

$\displaystyle \sf{ \implies \: v = 2 }$

From Equation 1 we get

$\displaystyle \sf{ u = - \frac{2}{4} = - 0.5\: }$

Let f = The focal length of the lens

Now we know that

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

$\displaystyle \sf{ \implies \: \frac{1}{f} = \frac{1}{ 2} - \frac{1}{ - 0.5} }$

$\displaystyle \sf{ \implies \: \frac{1}{f} = \frac{1}{ 2} + \frac{1}{ 0.5} }$

$\displaystyle \sf{ \implies \: \frac{1}{f} =0.5 + 2}$

$\displaystyle \sf{ \implies \: \frac{1}{f} = 2.5}$

$\displaystyle \sf{ \implies \: f = \frac{1}{2.5} }$

$\displaystyle \sf{ \implies \: f = 0.4}$

FINAL ANSWER

Hence the required focal length of the lens

= 0.4 cm

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Answer 2

Given:

An object is placed at a distance of 2.5 cm form a screen and a convex lens is interposed between them. The magnification produced is 4.

To find:

Focal length of lens?

Calculation:

• We will assume that image is formed on the screen, such that u + v = 2.5 cm.

Now, magnification be 'm'

• Now, for a convex lens magnified real image is always inverted, so magnification value is -4.

$\rm \: m = \dfrac{v}{u}$

$\rm \implies \: - 4 = \dfrac{v}{u}$

$\rm \implies \: v = - 4u$

Now, putting this value in 1st eq.

$\rm \implies \: |u| + |v| = 2.5$

$\rm \implies \: 4u + u = 2.5$

$\rm \implies \: 5u = 2.5$

$\rm \implies \:u = 0.5 \: cm$

Here sign of 'u' will be negative :

• So, u = -0.5 cm.

• So, v = -4u = 2 cm.

Applying Len's Law:

$\rm \: \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\rm \implies\: \dfrac{1}{f} = \dfrac{1}{2} - \dfrac{1}{( - 0.5)}$

$\rm \implies\: \dfrac{1}{f} = \dfrac{1}{2} + 2$

$\rm \implies\: \dfrac{1}{f} = \dfrac{5}{2}$

$\rm \implies\: f = 0.4 \: cm$

So, focal length of lens is +0.4 cm.

• + sign verifies that it is a CONVEX LENS (as stated in question).