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

18 c m

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15 c m

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20 c m

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25 c m

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Solution

The correct option is D $15 c m$
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: $u_{B} = - 15 c m$

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

The correct option is D 15 cmThe 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 cm

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