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If a lens of focal length 20 cm, made of glass of refractive index 1.5, is placed in a liquid of refractive index 1.25, the focal length of the lens becomes ___________. |
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Answer» Solution :`(1)/(20)=((15)/(1)-1)((1)/(R_(1))-(1)/(R_(2)))=0.5 ((1)/(R_(1))-(1)/(R_(2)))` and `(1)/(F)=((1.5)/(1.25)-1)((1)/(R_(1))-(1)/(R_(2)))=(0.25)/(1.25)((1)/(R_(1))-(1)/(R_(2))) RARR f=(0.5xx1.25)/(0.25)xx20=50cm` |
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