  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 lengthSpeed 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. Physics

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