Two objects A and B when placed in front of a concave mirror of focal length $7.5 c m$ , give images of equal size. If A is three times the size of B and is placed $30 c m$ from the mirror, what is the distance of B from the mirror?

10 c m

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12.5 c m

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15 c m

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17.5 c m

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Solution

The correct option is B $15 c m$
The relation between object distance, focal length and magnification is given by,

$u = f (1 + \frac{1}{m})$

Thus, $\frac{m_{A}}{m_{B}} = \frac{u_{B} - f}{u_{A} - f}$

or, $\frac{h_{B}}{h_{A}} = \frac{u_{B} - f}{u_{A} - f}$

or, $\frac{1}{3} = \frac{u_{B} + 7.5}{- 30 + 7.5}$

Solving this, we get: $u_{B} = - 15 c m$

So the distance of $B$ from the mirror is $15 c m$

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Two objects A and B when placed in front of a concave mirror of focal length 7.5cm, give images of equal size. If A is three times the size of B and is placed 30cm from the mirror, what is the distance of B from the mirror?

The correct option is B 15cmThe relation between object distance, focal length and magnification is given by,u=f(1+1m)Thus, mAmB=uB−fuA−for, hBhA=uB−fuA−for, 13=uB+7.5−30+7.5Solving this, we get: uB=−15 cmSo the distance of B from the mirror is 15cm

Two objects A and B when placed in front of a concave mirror of focal length $7.5 c m$ , give images of equal size. If A is three times the size of B and is placed $30 c m$ from the mirror, what is the distance of B from the mirror?