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A convex lens of refractive index 3/2 has a power of `2.5^(@)`. If it is placed in a liqud of refractive index 2,the new power of the lens isA. 2.5DB. `-2.5D`C. 1.25DD. `-1.25D` |
Answer» Correct Answer - D (d) We have,P=`(µ-1)[1/R_(1)-1/R_(2)]` `2.5=(3/2-1)(1/R_(1)-1/R_(2))` `P=((3//2)/2-1)(1/R_(1)-1/R_(2))` `therefore` P=-1.25D` |
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