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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. |
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Answer» `62.2cm` `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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