Question

a source of light is at 10cm from a convex mirror and is then moved to a distance of 2cm from the mirror how much does the image move if the radiu of curvature of the mirror is 4.8cm

Answer 1

And is 0.8cm :Find v1 and v2 and find there difference. Also Focal length will be same for both u1 and u2

Answer 2

The image move around 0.8 cm

Solution:

The given question suggest that the source of light is kept in front of a convex mirror and is being moved then the radius of curvature is given accordingly, the image moves to a distance as calculated as follows,

Given:  $u_{1}$= - 10 cm   $u_{2}$ = - 2cm  R = 4.8 cm

$\begin{array}{l}{f=\frac{R}{2}} \\ \\{f=\frac{4.8}{2}}\end{array}$

f = 2.4 cm

As we know that $\frac{1}{v_{1}}=\frac{1}{f}-\frac{1}{u_{1}}$

$\begin{array}{l}{=\frac{1}{2.4}-\frac{1}{-10}=\frac{10}{24}+\frac{1}{10}} \\ \\{=\frac{5}{12}+\frac{1}{10}=\frac{25+6}{60}=\frac{31}{60}} \\ \\{\frac{1}{v_{1}}=\frac{31}{60}} \\ \\{v_{1}=\frac{60}{31}=1.935 \mathrm{cm}}\end{array}$

When $u_{2}=-2 \mathrm{cm}$

$\begin{array}{l}{\frac{1}{v_{2}}=\frac{1}{f}-\frac{1}{u_{2}}} \\ \\{\frac{1}{v_{2}}=\frac{1}{2.4}-\frac{1}{-2}=\frac{1}{2.4}+\frac{1}{2}=\frac{10}{24}+\frac{1}{2}=\frac{5}{12}+\frac{1}{2}=\frac{5+6}{12}=\frac{11}{12}}\end{array}$

$\frac{1}{v_{2}}=\frac{12}{11}=1.09 \mathrm{cm}$

Therefore $\bold{v_{2}-v_{1}}$ = 1.9 – 1.1 = 0.8 cm.