Question

Draw a ray diagram for a concave mirror and prove that focal length of the mirror is half of its radius of curvature. Write mirror equation for a concave mirror. When light enters from rare to denser medium, what effect occurs on the values of wavelength and frequency of it?

Answer 1

In the given figure a ray of light BP' traveling parallel to the principal axis PC is incident on a spherical mirror PP'. The distance PF is equal to the focal length $f$. C is the centre of curvature. The distance PC is equal to the radius of curvature R of the mirror. P'C is the normal to the mirror at the point of incidence P'. In the given figure a ray of light BP' travelling parallel to the principal axis PC is incident on a spherical mirror PP'. After reflection, it goes along P'R, through the focus F.

For concave mirror,

$\angle B{P}^{\prime }C=\angle P^{\prime }CF$ (alternate angle)

$\angle B{P}^{\prime }C=\angle C{P}^{\prime }F$ (law of reflection $\angle i=\angle r$)

Hence $\angle P^{\prime }CF=\angle C{P}^{\prime }F$

$\therefore △F{P}^{\prime }C$ is isosceles.

Hence $P^{\prime }F=PF$

If aperture is small ,point $P^{\prime }$ is close to point $P$

then, $P^{\prime }F=PF$

$\therefore PF=FC=\frac{1}{2}PC$

$f=\frac{1}{2}R$

According to mirror equation of concave mirror,

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

where:

$u:$ Object distance

$v:$ Image distance

$f:$ Focal length

Speed of light is higher in rarer medium.

$v_2<v_1$

Frequency of light does not change on change of medium.

$f_1=f_2$

$\frac{v_1}{{\lambda }_1}=\frac{v_2}{{\lambda }_2}$

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{v_2}{v_1}<1$

Hence, when light enters from rarer to denser medium, wavelength reduces but frequency is unchanged.