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A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. Calculate the magnification |
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Answer» Solution :longitudinal magnification (m) = `("lenght of image"(L.))/("lenght of object"(l))` Give: length of object, l = `(f)/(3)` LET, l be the length of image, then, `m=(l.)/(l)=(l.)/(f//3)(or)l=(mf)/(3)` Image of one end coincides with the object. THUS, the coinciding end must be at center of curvature. Hence, U = R = 2f `u = u + (f)/(3)` `u=u.(f)/(3)=2f-(f)/(3)=(5f)/(3)` `v=u+(f)/(3)+(mf)/(3)=(5f)/(3)+(f)/(3)+(mf)/(3)=(f(6+m))/(3)` Mirror equation, `(1)/(v) + (1)/(u) = (1)/(f)` `(1)/(-((f(6+m))/(3)))+(1)/(-((5f)/(3)))=(1)/(-f)` After equation, `(3)/(f(6+m))+(3)/(5f)=(1)/(f),(3)/((6+m))=(2)/(5)` `6 + m = (15)/(2) , m = (15)/(2) - 6` `m = (3)/(2) = 1.5` 
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