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Question
The focal length of a this lens in vacuum is f. If the material of the lens has µ = 3/2, its focal length when immersed in water refractive index 4/3 will
Options
F
4f/3
3f
4f
MCQ
Solution
4f
Explanation:
`1/(Fa) = ((ug)/(ua) - 1) (1/R_1 - 1/R_2)`
`1/(Fw) = ((ug)/(muw) - 1) (1/R_1 - 1/R_2)`
∴ Dividing (i) by (ii) we get
`(fw)/(fa) = (((mug)/(mua - 1)))/(((mug)/(muw - 1))`
= `((3/(2 - 1)))/(((3/2)/(4/2) = 1)`
= `(1/2)/(1/0)` = 4
fw = 4fa = 4f
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Diffraction of Light - The Single Slit
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