A concave lens of glass, refractive index 1.5, has both surface of the same radius of curvature R. On immersion in a medium of refractive index 1.75, it will

Asked by yashu.fly | 22nd Nov, 2014, 10:18: PM

Expert Answer:

Given that:
Refractive index of glass lens, n2 = 1.5 ,
Refractive index of the medium in which the lens is immersed, n1 =1.75
Let f be the focal length of the lens.
begin mathsize 14px style By space lens space makers space formula comma space we space know space that : 1 over straight f equals open parentheses straight n subscript 2 over straight n subscript 1 minus 1 close parentheses open parentheses 1 over straight R subscript 1 minus 1 over straight R subscript 2 close parentheses We space know space that space for space straight a space concave space lens space straight R subscript 1 space is space plus ve space and space straight R subscript 2 space is space minus space ve Also space we space know space that space focal space length space of space concave space lens space is space minus space ve Here space given space that space straight R subscript 1 equals plus straight R space and space straight R subscript 2 equals minus straight R Substituting space the space values space we space get : fraction numerator 1 over denominator minus straight f end fraction equals open parentheses fraction numerator 1.5 over denominator 1.75 end fraction minus 1 close parentheses open parentheses 1 over straight R minus fraction numerator 1 over denominator minus straight R subscript blank end fraction close parentheses fraction numerator 1 over denominator minus straight f end fraction equals open parentheses fraction numerator 1.5 minus 1.75 over denominator 1.75 end fraction close parentheses open parentheses 2 over straight R close parentheses fraction numerator 1 over denominator minus straight f end fraction equals open parentheses fraction numerator minus 0.25 over denominator 1.75 end fraction close parentheses open parentheses 2 over straight R close parentheses fraction numerator 1 over denominator minus straight f end fraction equals fraction numerator minus 0.5 over denominator 1.75 straight R end fraction fraction numerator 1 over denominator minus straight f end fraction equals fraction numerator 1 over denominator minus 3.5 straight R end fraction rightwards double arrow straight f equals 3.5 straight R rightwards double arrow Focal space length space becomes space positive space. straight i. straight e. space it space behaves space like space straight a space convex space lens That space is space inside space the space medium space the space lens space will space behave space like space straight a space converging space lens space of space focal space length space 3.5 space straight R. end style
 

Answered by Jyothi Nair | 22nd Nov, 2014, 11:07: PM