Question

Find the co-ordinates of image of point object P formed after two successive reflection at concave mirror and then at convex.

Answer 1

To find the co-ordinates of image of point object P formed after two successive reflection at concave mirror and then at convex.

Solution:

Consider concave mirror ($M_1$) from figure,

Object distance, $u_1=-20cm$

Focal length of the mirror, $f_1=-15cm$

Applying mirror formula,

$\frac{1}{f_1}=\frac{1}{v_1}+\frac{1}{u_1}⟹\frac{1}{v_1}=\frac{1}{f_1}-\frac{1}{u_1}⟹\frac{1}{v_1}=\frac{1}{-15}-\frac{1}{-20}⟹\frac{1}{v_1}=\frac{-4+3}{60}⟹v_1=-60cm$

Magnification, $m_1=-\frac{v_1}{u_1}=-\frac{(-60)}{(-20)}=-3$ (image is inverted)

The height at which the image point will be $m_1\times AP=3\times 2=6mm$

The image formed by concave mirror acts as object for convex mirror.

Now, at convex mirror ($M_2$) from figure

Object distance, $u_2=+10cm$

Focal length of the mirror, $f_2=20cm$

Applying mirror formula, we get

$\frac{1}{f_2}=\frac{1}{v_2}+\frac{1}{u_2}⟹\frac{1}{v_2}=\frac{1}{f_2}-\frac{1}{u_2}⟹\frac{1}{v_2}=\frac{1}{20}-\frac{1}{10}⟹\frac{1}{v_2}=\frac{1-2}{20}⟹v_2=-20cm$

Hence the final position of image after two successive reflection at concave mirror and then at convex 20 cm in front of convex mirror, $M_2$

Magnification $m_2=-\frac{v_2}{u_2}=-\frac{(-20)}{10}=+2$

So the image of the point will be formed above, $m_2\times C^{\prime }{P}^{\prime }=2\times 8=16mm$

So the required coordinates of the image is, $P(30,-14)$