When an object is placed at a distance of 25 | vedclass

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(D) The magnification $m$ for a spherical mirror is given by $m = \frac{f}{f-u}$.For the first position,$u_1 = -25 \, cm$,so $m_1 = \frac{f}{f - (-25)

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) The magnification $m$ for a spherical mirror is given by $m = \frac{f}{f-u}$.For the first position,$u_1 = -25 \, cm$,so $m_1 = \frac{f}{f - (-25)} = \frac{f}{f+25}$.The object is moved $15 \, cm$ farther away,so the new position is $u_2 = -(25 + 15) = -40 \, cm$.The magnification at the second position is $m_2 = \frac{f}{f - (-40)} = \frac{f}{f+40}$.Given the ratio $\frac{m_1}{m_2} = 4$,we substitute the expressions:$\frac{f/(f+25)}{f/(f+40)} = 4$$\frac{f+40}{f+25} = 4$$f + 40 = 4(f + 25)$$f + 40 = 4f + 100$$3f = -60$$f = -20 \, cm$.The magnitude of the focal length is $20 \, cm$.

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