The plane faces of two identical plano-convex | vedclass

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(B) When two identical plano-convex lenses are placed with their plane faces together,they form a biconvex lens.Given,focal length of each plano-conve

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) When two identical plano-convex lenses are placed with their plane faces together,they form a biconvex lens.Given,focal length of each plano-convex lens,$f_1 = f_2 = 40 \, cm$.The power of each lens is $P_1 = P_2 = \frac{100}{f} = \frac{100}{40} = 2.5 \, D$.The power of the combination is $P = P_1 + P_2 = 2.5 + 2.5 = 5 \, D$.The focal length of the combination is $F = \frac{100}{P} = \frac{100}{5} = 20 \, cm$.For a real,inverted image with magnification $m = -1$,the object must be placed at a distance of $2F$ from the lens.Therefore,the object distance $u = 2 	imes 20 \, cm = 40 \, cm$.

The plane faces of two identical plano-convex lenses,each having a focal length of $40 \, cm$,are placed against each other to form a biconvex lens. The distance from this lens at which an object must be placed to obtain a real,inverted image with magnification $-1$ is ...... $cm$.

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