Question

For a given lens , the magnification was found to be twice as large as when the object was 0.15m distant it as when the distance was 0.2 m the focal length of the lens is​

Answer 1

ANSWER:

• Focal length of the lens = 10 cm.

GIVEN:

• For a given lens, the magnification was found to be twice as large as when the object was 0.15 m distant as it when the distance was 0.2 m.

TO FIND:

• The focal length of the lens.

EXPLANATION:

$\bigstar\boxed{ \bold{ \large{ \blue{ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} }}}} \\ \\ \\ \sf \dfrac{1}{f} = \dfrac{1}{v_1 } - \dfrac{1}{ - u_1} \\ \\ \\ \sf Multiply \ by \ u_1\ on \ both \ sides.\\ \\ \\ \sf \dfrac{u_1}{f} = \dfrac{u_1}{v_1 } + \dfrac{ u_1}{u_1} \\ \\ \\ \bigstar\boxed{ \bold{ \large{ \orange{ m = - \dfrac{v}{u} }}}}\\ \\ \\ \sf \dfrac{u_1}{f} = - \dfrac{1}{m_1}+ 1$

$\sf \dfrac{u_1}{f} - 1 = - \dfrac{1}{m_1} \\ \\ \\ \sf \dfrac{u_1 - f}{f} = - \dfrac{1}{m_1} \\ \\ \\ \sf m_1= \frac{f}{f -u_1 } \\ \\ \\ \sf Similarly \ m_2= \frac{f}{f -u_2 } \\ \\ \\ \sf 2m_2 = m_1 \\ \\ \\ \sf2 \times \frac{f}{f -u_2 } = \frac{f}{f -u_1 } \\ \\ \\ \sf 2(f -u_1) = f -u_2 \\ \\ \\ \sf f = 2u_1 - u_2 \\ \\ \\ \sf u_1 = 0.15 \ m, \ u_2 = 0.2 \ m. \\ \\ \\ \sf f = 2(0.15) - 0.2 \\ \\ \\ \sf f = 0.3 - 0.2 = 0.1 \ m$

Hence the focal length of the lens = 10 cm.

Answer 2

Given :-

For a given lens , the magnification was found to be twice as large as when the object was 0.15m distant it as when the distance was 0.2 m

To Find :-

Focal length

Solution :-

We know that

m = -v/u

M = 2M

By using lens formula

1/u + 1/v = 1/f

1/v = 1/f - 1/u

v = 1(u) - 1(f)/uf

v = u - f/uf

So, Now

f = 2u - u'

f = 2(0.15) - 0.2

f = 0.3 - 0.2

f = 0.1 m

1 m = 100 cm

0.1 m = 10 cm

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