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O is a point object kept on the principal axis of a concave mirror M of radius of curvature 20cm. P is a prism of angle 1.8^@ . Light falling on the prism (at small angle of incidence) gets refracted through the prism and then falls on the mirror. Refractive index of the prism is 3//2 . Find the distance between theimages formed by the concave mirror due to this light. |
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Answer» `=(1.5-1)1.8^(@)` `=(0.5)1.8xx(pi)/(180)rad=(pi)/(200)rad` `:.` Distance `OO_(1)=10xx(pi)/(200)=(pi)/(20)cm` Similarly, distance `OO_(2)=(pi)/(20)cm` `:. O_(1)O_(2)=(pi)/(10)cm` For the mirror, `O_(1)` and `O_(2)` are two point OBJECTS, at a distance of30cm. Now, applying mirror FORMULA `(1)/(v)+(1)/(-30)=(1)/(-10)rArrv=-15cm` Latent magnificationis `-(v)/(u)=((-15))/(-30)=-(1)/(2)` `(-ve` sign indicates INVERTED image `)` Therefore, if `O_(1)^(')` and`O_(2)^(')` are the images formed, then distance between them, `O_(1)^(')O_(2)^(')=(1)/(2)O_(1)O_(2)=(1)/(2).(pi)/(10)=(pi)/(20)cm` 
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