1.

Find the change in the focal lengthof the lens, if a convex lens of focal length 20cm and refractive index 1.5, is immersed in water having refractive index 1.33.

Answer»

`62.2cm`
`5.82cm`
`58.2cm`
`6.22cm`

Solution :(c) When the lens is in AIR `1/(f_(a))=(mu-1)(1/(R_(1))-1/(R_(2)))`
`1/20=(1.5-1)(1/(R_(1))-1/(R_(2)))` or `(1/(R_(1))-1/(R_(2)))=1/10`
When lens is in water
`1/(f_(w))=((mu_(G)-mu_(w))/(mu_(w)))(1/(R_(1))-1/(R_(2)))=((1.5-1.33)/1.33)xx1/10`
`f_(w)=78.2cm`
The change in FOCAL length`=78.2-20=58.2cm`


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