Question

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

• A. -15 cm
• B. 25 cm
• C. -17.5 cm
• D. -11 cm

Answer 1

The correct option is C -17.5 cm

Given,

The size of the image of object A and B is the same, hence

${h_A}^{\prime }={h_B}^{\prime }$

The size of object A is four times that of B, hence

$h_A=3h_B$

Now the magnification of object A can be given as

$m_A=\frac{{h_A}^{\prime }}{h_A}$ …… (1)

Similarly for object B

$m_B=\frac{{h_B}^{\prime }}{h_B}$ ……. (2)

From equation (1) and (2) the ratio of magnification is given as

$\frac{m_A}{m_B}=\frac{{h_A}^{\prime }}{h_A}\times \frac{h_B}{{h_B}^{\prime }}\Rightarrow \frac{m_A}{m_B}=\frac{{h_A}^{\prime }}{3h_B}\times \frac{h_B}{{h_A}^{\prime }}$

$\therefore \frac{m_A}{m_B}=\frac{1}{3}$ ……. (3)

Now the magnification formula is given as

$m=\frac{f}{f-u}$

Therefore equation (3) becomes

$\frac{m_A}{m_B}=\frac{f-u_B}{f-u_A}\Rightarrow \frac{1}{3}=\frac{-7.5-u_B}{-7.5+30}$

$-u_B=\frac{22.5}{3}+10\Rightarrow u_B=-17.5cm$

Hence the object B is placed at $-17.5cm$ from the mirror.