A concave mirror forms a real image of an ...

CBSE Class 12 Physics 2020 Outside Delhi Set 3 Paper

Question : 12 of 13

Marks: +1, -0

A concave mirror forms a real image of an object kept at a distance

9\ \mathrm{cm}

from it. If the object is taken away from the mirror by

6\ \mathrm{cm}

, the image size reduces to

\frac{1}{4}

th of its previous size. Find the focal length of the mirror.

Solution:

For the

1^{\text{st}}

position :

\text{ Object distance } = u_1 = -9\ \mathrm{cm}

\text{ Image size } = h_1'

\text{ Object size } = h_1

Using mirror formula

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

\text{ Or, } -\frac{1}{9} + \frac{1}{v_1} = \frac{1}{f}

.........(i)

\text{ Magnification } = \frac{h_1'}{h_1} = \frac{v_1}{u_1}

For the

2^{\text{nd}}

position :

\text{ Object distance } = u_2 = -15\ \mathrm{cm}

\text{ Image size } = h_2' = \frac{1}{4} h_1'

\text{ Object size } = h_1

Using mirror formula

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

\text{ Or, } -\frac{1}{15} + \frac{1}{v_2} = \frac{1}{f}

........(ii)

\text{ Magnification } = \frac{h_2'}{h_1} = \frac{\frac{1}{4} h_1'}{h_1} = \frac{v_2}{u_2}

\text{ Or, } \frac{1}{4} \frac{v_1}{u_1} = \frac{v_2}{u_2}

\text{ Or, } v_2 = \frac{5}{12} v_1

\text{ Comparing equations (i) and (ii), }

-\frac{1}{9} + \frac{1}{v_1} = -\frac{1}{15} + \frac{1}{v_2}

\text{ Or, } \frac{1}{15} - \frac{1}{9} = \frac{1}{v_2} - \frac{1}{v_1}

\text{ Or, } -\frac{2}{45} = \frac{12}{5 v_1} - \frac{1}{v_1}

\text{ Or, } -\frac{2}{45} = \frac{7}{5 v_1}

\therefore v_1 = -\frac{63}{2}\ \mathrm{cm}

Putting in eqn (i),

-\frac{1}{9} - \frac{2}{63} = \frac{1}{f}

\text{ Or, } -\frac{1}{7} = \frac{1}{f}

\therefore f = \text{ focal length } = -7\ \mathrm{cm}

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