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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. 2.5 DB. `-2.5 D`C. `1.25 D`D. `- 1.25 D` |
Answer» Correct Answer - D We have, `P=(mu-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=-1.25 D` |
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