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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?


Solution

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,
$$\angle BP'C = \angle P'CF$$ (alternate angle)
$$\angle BP'C=\angle CP'F $$ (law of reflection $$\angle i = \angle r$$)
Hence $$\angle P'CF=\angle CP'F$$
$$\therefore \triangle FP'C$$ is isosceles.
Hence $$P'F=PF$$
If aperture is small ,point $$P'$$ is close to point $$P$$
then, $$P'F=PF$$
$$\therefore\ PF=FC=\dfrac{1}{2}PC$$
$$f=\dfrac{1}{2}R$$

According to mirror equation of concave mirror,
$$\cfrac{1}{u} + \cfrac{1}{v} = \cfrac{1}{f}$$
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.
$$f_1 = f_2$$
$$\cfrac{v_1}{\lambda_1} = \cfrac{v_2}{\lambda_2}$$
$$\cfrac{\lambda_2}{\lambda_1} = \cfrac{v_2}{v_1} <1$$
Hence, when light enters from rarer to denser medium, wavelength reduces but frequency is unchanged.

631508_604136_ans_225a0d4692c846b8ba4c324b1238302e.png

Physics

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