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the power of convex lens of refractive index 1.5 is 0.5D in air in water refractive index 4÷3 it's focal length will be

Asked by racchurashmi74 | 10 Feb, 2021, 06:05: AM

Expert Answer

We have Lens makers formula as

begin mathsize 16px style 1 over f subscript a space equals space open parentheses mu presuperscript a subscript g space minus space 1 close parentheses space open parentheses 1 over R subscript 1 minus 1 over R subscript 2 close parentheses space end style

............................(1)

where f_a is focal length of lens in air , ^aμ_g is refractive index of glass with respect to air , R₁and R₂ are radii of curvature

We are given that power of lens in air is 0.5D , i.e if focal length is expressed in metre , ( 1/ f_a ) = 0.5 or f_a = 2 m

Using (1/f_a) , we rewrite eqn.(1) as

begin mathsize 16px style space open parentheses mu presuperscript a subscript g space minus space 1 close parentheses space open parentheses 1 over R subscript 1 minus 1 over R subscript 2 close parentheses space equals space 0.5 end style

.......................(2)

Let power of lens in water be x. Then using lens makers formula in water , we get an equation similar to eqn.(5) as

begin mathsize 16px style space open parentheses mu presuperscript w subscript g space minus space 1 close parentheses space open parentheses 1 over R subscript 1 minus 1 over R subscript 2 close parentheses space equals space x end style

...........................(3)

where ^wμ_g is refractive index of glass with respect to air

we get ^wμ_g=^aμ_g / ^aμ_w = (3/2) / (4/3) = 9/8 = 1.125

where ^aμ_w is refractive index of water with respect to air

Dividing eqn.(3) by eqn.(2) , we get ( x / 0.5 ) = ( ^wμ_g - 1 ) / ( ^aμ_g - 1 ) = ( 1.125 -1 ) / ( 1.5 -1 )

Hence we get x = 0.125 D or focal length of lens in water , f_w = 1 / 0.125 = 8 m

Answered by Thiyagarajan K | 10 Feb, 2021, 10:42: AM

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