Question

A source of light is at 10 cm from a convex mirror and is then moved to a distance of 2 cm from the mirror. How much does the image move, if the radius of curvature of the mirror is 4.8 cm?

• A. 0.95 cm
• B. 0.85 cm
• C. 0.75 cm
• D. 0.65 cm

Answer 1

The correct option is B 0.85 cm

$u=-10cm$

$u_2=-2cm$ $R_{2}u-8$

$f=\frac{R}{2}=\frac{2u-8}{2}=2-u$

By $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\Rightarrow \frac{1}{v_1}=\frac{1}{f}-\frac{1}{u_1}$

$=\frac{1}{2-u}-\frac{1}{u_1}=\frac{31}{60}$

$=\frac{1}{2.u}-\left(\frac{-1}{10}\right)=\frac{5}{12}+\frac{1}{10}=\frac{31}{60}$

$\therefore v_1=\frac{60}{31}=1.930$

when $u_2=-2cm\frac{1}{v_2}=\frac{1}{f}-\frac{1}{u_2}=\frac{1}{10}$

$=\frac{10}{2u}-(-1/2)=\frac{5}{12}+\frac{1}{2}$

$\therefore v_2=1.09cm$

$v_1-v_2=1.935-1.09=1.935-1.1$

$=0.8cm$