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Flint glass lens is made of refractive index 1.5. When it is placed in a liquid of refractive index 1.25, then its focal length will be ....

Answer»

1.25 f
2.5 f
1.2 f
1.3 f

Solution :Let, FOCAL LENGTH is f,
`therefore 1/f=(""_amu_g-1)((1)/(R_1)-(1)/(R_2))`
`=(1.5-1)((1)/(R_1)-(1)/(R_2))`
`1/f=0.5((1)/(R_1)-(1)/(R_2))`
`therefore (1)/(0.5 f)=(1)/(R_1)-(1)/(R_2)`…...(1)
Let, after INSERTING it in LENS, it BECOMES f.,
`therefore (1)/(f)=(""_lmu_g-1)((1)/(R_1)-(1)/(R_2))`
`((1.5)/(1.25)-1)((1)/(R_1)-(1)/(R_2))`
`(1)/(f.)=((0.25)/(1.25))xx(1)/(0.5f)`
`therefore(1)/(f.)=(1)/(2.5f)`
`therefore f. =2.5` f


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