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Aconvex mirror and a concave mirror each of focal length 10 cm are placed coaxially. The are separated by40 cm and their reflecting surfaces face each other. A point object is kept on the principle axis at a distance x cm from the concave mirror such that final image after two reflections, first on the concave mirror, is on the object itself. Find integer next to x . |
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Answer» `U=-x` `f=-f` `(1)/(V)=(1)/(f)-(1)/(u)=-(1)/(f)+(1)/(x)rArrv=(fx)/(f-x)` `therefore` Distance of the image from the convex mirror `=4f+(fx)/(f-x)=(4f^(2)-3fx)/((f-x))` `u=-((4f^(2)-3fx))/((f-x))` `f=f` By the question, `v=-(4f-x)``therefore``(1)/(v)=(1)/(f)-(1)/(u) gives` , `(1)/(-(4f-x))=(1)/(f)+((f-x))/((4f^(2)-3fx))` on solving, we GET, `x^(2)-6fx+6f^(2)=0` which gives, `x=(3+-sqrt3)f` But `xlt4f` Hence, `x=(3-sqrt3)10 cm=12.68 cm` 
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