Question

Two objects A and B when placed one after another infront of a concave mirror of focal length 10 cm form images of same size. Size of object A is four times that of B. If object A is placed at a distance of 50 cm from the mirror, what should be the distance of B from the mirror?

• A. 10 cm
• B. 20 cm
• C. 30 cm
• D. 40 cm

Answer 1

Answer: (B)

Step 1: Given that:

Two objects A and B are placed in front of a concave mirror.

The focal length(f) of the concave mirror = $-10cm$

Size of the image of object A ($h_{i_A}$)= Size of the image of object B($h_{i_B}$)

Size of object A = $4\times$ size of object B

$(h_{o_A}=4h_{o_B})$

Object distance(u) for object A = $-50cm$

(Note: The focal length of the concave mirror and the object distance are taken negative as they lie left to the pole of the mirror)

Step 2: Calculation of the object distance for object B:

From the mirror formula,

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

1f=1v+1u\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
and magnification formula,

$m=\frac{h_i}{h_o}=-\frac{v}{u}=\frac{f}{f-u}$

Where; v is the image distance, u= object distance and f is the focal length of the mirror.

Now, for object A, we have;

$m_A=\frac{h_{i_A}}{h_{o_A}}=\frac{-10cm}{-10-(-50cm)}$

$m_A=\frac{h_{i_A}}{h_{o_A}}=\frac{-10}{-10+50}$

$m_A=\frac{h_{i_A}}{h_{o_A}}=\frac{-10}{40}$

$m_A=\frac{h_{i_A}}{h_{o_A}}=\frac{-1}{4}$ ..........(1)

For object B;

$m_B=\frac{h_{i_B}}{h_{o_B}}=\frac{-10}{-10cm-u_B}$

$m_B=\frac{h_{i_B}}{h_{o_B}}=\frac{-10}{-10-u}$........(2)

Now, using the relation $(h_{o_A}=4h_{o_B})$ and $h_{i_A}$ = $h_{i_B}$ , we get from equation 1)

$m_A=\frac{h_{i_B}}{\text{4}{h}_{o_B}}=\frac{-1}{4}$

$\frac{h_{i_B}}{h_{o_B}}=\frac{-1}{1}$

Thus, from equation 2) we get;

$\frac{-1}{1}=\frac{-10}{-10-u_B}$

$-10=10+u_B$

$10+u=-10$

$u_B=-20cm$

Thus,

Object B should be placed at 20 cm in front of the concave mirror.

m=hiho=−vum=\frac{h_i}{h_o}=-\fra