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A double convex lens made of glass refractive index 1.6 has its both surfaces of equal radii of curvature of 30 cm each. An object of 5 cm height is placed at a distance of 12.5 cm from the lens. Find the position, nature and size of the image. |
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Answer» Solution :For double convex lens `R_(1) = +30 cm` and `R_(2) = -30 cm, n=1.6, u =-12.5 cm` Using relatino `1/f=(n-1)(1/R_(1)+1/R_(2))`, we have `1/f =(1.6-1) (1/30 + 1/30)= 0.6 xx 2/30 RARR f= 30/(0.6 xx 2) = 25 cm` Using lens formula `1/v -1/u = 1/f`, we have `1/v =1/f+1/u = 1/25 + 1/(-12.5) = (1-2)/(25) = -1/25 rArr v=-25 cm` Height of object h = 5 cm. If height IMAGE be `h.`, then `h^(.)/h = v/u` `therefore h. = v/u.h = (-25)/(-12.5) xx 5 = 10 cm` Thus, a virtual and erect image of 10 cm height is formed at 25 cm from the lens. |
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