The two surfaces of a biconvex lens has same radii of curvatures. This lens is made of glass of refractive index 1 . 5 and has a focal length of 10 cm in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. The two pieces are glued such that the convex surfaces touch each other. If this combination lens is immersed in water (refractive index = 4 3 ), its focal length (in cm ) is

Question

The two surfaces of a biconvex lens has same radii of curvatures. This lens is made of glass of refractive index 1 . 5 and has a focal length of 10 cm in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. The two pieces are glued such that the convex surfaces touch each other. If this combination lens is immersed in water (refractive index = 4 3 ), its focal length (in cm ) is, 5 cm, 10 cm, 20 cm, 40 cm

MARKS App · Accepted Answer

If a lens of focal length f is divided into two equal parts and each has a focal length f ′ then 1 f = 1 f ′ + 1 f ′ , i.e, f ′ = 2 f Now if these parts are put in contact, then resultant focal length of the combination will be 1 F = 1 2 f + 1 2 f ,i.e, F = f (initial value) For this combination, 1 F = a μ g - 1 1 R 1 - 1 R 2 … … … … . . ( i ) Now, if this combination is immersed in liquid, then 1 F ′ = 1 μ g - 1 1 R 1 - 1 R 2 … . . ( i i ) F ′ f = ( a μ g - 1 ) ( 1 μ g - 1 ) = 1 .5 - 1 3 2 4 3 - 1 o r F ′ f = 0.5 9 8 - 1 = 0.5 × 8 ∴ F ' = 0.5 × 8 × 10 = 40 cm ∴ F ' = 0.5 × 8 × 10 = 40 cm

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