A convex lens of focal length 20cm and made of glass of refractive index 1.5 is immersed in water of reffractive index 1.33 calculate change in focal length of the lens 

Asked by Sandeep Yadav | 6th Dec, 2014, 11:09: PM

Expert Answer:

begin mathsize 14px style bold Case bold space bold 1 bold : bold Lens bold space bold in bold space bold air Let space the space focal space length space of space lens space in space air space be space straight f subscript air space Given space that : straight f subscript air space equals 20 space cm comma space straight n subscript 1 space equals 1 space left parenthesis air right parenthesis space comma space straight n subscript 2 subscript space end subscript equals 1.5 space left parenthesis glass right parenthesis According space to space lens space makers space formula : 1 over straight f subscript air equals open square brackets straight n subscript 2 over straight n subscript 1 minus 1 close square brackets open square brackets 1 over straight R subscript 1 minus 1 over straight R subscript 2 close square brackets  1 over 20 equals open square brackets fraction numerator 1.5 over denominator 1 end fraction minus 1 close square brackets open square brackets 1 over straight R subscript 1 minus 1 over straight R subscript 2 close square brackets 1 over 20 equals 0.5 space open square brackets 1 over straight R subscript 1 minus 1 over straight R subscript 2 close square brackets space minus minus minus minus minus minus minus minus minus left parenthesis 1 right parenthesis bold Case bold space bold 2 bold : bold space bold Lens bold space bold in bold space bold water Let space the space focal space length space of space lens space in space water space be space straight f subscript water Given space that : space straight n subscript 1 space equals 1.33 space left parenthesis water right parenthesis space comma space straight n subscript 2 subscript space end subscript equals 1.5 space left parenthesis glass right parenthesis straight f subscript water equals space ? According space to space lens space makers space formula : 1 over straight f subscript water equals open square brackets straight n subscript 2 over straight n subscript 1 minus 1 close square brackets open square brackets 1 over straight R subscript 1 minus 1 over straight R subscript 2 close square brackets  1 over straight f subscript water equals open square brackets fraction numerator 1.5 over denominator 1.33 end fraction minus 1 close square brackets open square brackets 1 over straight R subscript 1 minus 1 over straight R subscript 2 close square brackets 1 over straight f subscript water equals 0.128 space open square brackets 1 over straight R subscript 1 minus 1 over straight R subscript 2 close square brackets space minus minus minus minus minus minus minus minus minus left parenthesis 2 right parenthesis  Dividing space left parenthesis 1 right parenthesis space by space left parenthesis 2 right parenthesis space we space get : straight f subscript water over 20 equals fraction numerator 0.5 over denominator 0.128 end fraction rightwards double arrow straight f subscript water space equals fraction numerator 0.5 over denominator 0.128 end fraction cross times 20 rightwards double arrow straight f subscript water space equals 78.125 space cm straight i. straight e space focal space length space in space water space equals space 78.125 space cm Therefore space change space in space focal space length space of space the space lens space when space it space is space immersed space in space water space equals space 78.125 space cm space minus space 20 space cm space equals 58.125 space cm end style

Answered by Jyothi Nair | 7th Dec, 2014, 08:50: PM