A thin convex lens L (refractive index = 1.5 ) is placed on a plane mirror M . When a pin is placed at A , such that O A = 18 c m , its real inverted image is formed at A itself, as shown in figure. When liquid of refractive index μ l is put between the lens and the mirror, the pin has to be moved to A ' , such that O A ' = 27 c m , to get its inverted real image at A ' itself. The value of μ l will be

Question

A thin convex lens L (refractive index = 1.5 ) is placed on a plane mirror M . When a pin is placed at A , such that O A = 18 c m , its real inverted image is formed at A itself, as shown in figure. When liquid of refractive index μ l is put between the lens and the mirror, the pin has to be moved to A ' , such that O A ' = 27 c m , to get its inverted real image at A ' itself. The value of μ l will be, 4 3, 3, 3 2, 2

MARKS App · Accepted Answer

1 f 1 = 1 2 × 2 18 1 f 2 = μ 1 - 1 - 18 When μ 1 is filled between lens and mirror, P = 2 18 - 2 18 μ 1 - 1 = 2 - 2 μ 1 + 2 18 ⇒ F m = - 18 2 - μ 1 2 = 6 - 3 μ 1 3 μ 1 = 4 μ 1 = 4 3

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