Ray Optics

JEE Main level, numerical answer, medium

173 of 295

Question 173 of 295

A convex lens of refractive index 1.51.5 and focal length 18cm18cm in air is immersed in water. The change in focal length of the lens will be_____cmcm.
(Given refractive index of water=43=\dfrac{4}{3})

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Answer: 54

From Lens Maker's formula,

For air: 1fa=(μgμa−1)(1R1−1R2)\dfrac{1}{{f}_{a}}=\left(\dfrac{{\mu }_{g}}{{\mu }_{a}}-1\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)

For water: 1fw=(μgμw−1)(1R1−1R2)\dfrac{1}{{f}_{w}}=\left(\dfrac{{\mu }_{g}}{{\mu }_{w}}-1\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)

∴1fa1fw=(μgμa−1)(μgμw−1)∴\dfrac{\dfrac{1}{{f}_{a}}}{\dfrac{1}{{f}_{w}}}=\dfrac{\left(\dfrac{{\mu }_{g}}{{\mu }_{a}}-1\right)}{\left(\dfrac{{\mu }_{g}}{{\mu }_{w}}-1\right)}
⇒fwfa=(1.5−1)(1.5×34−1)⇒fw=0.50.125×18=72cm\Rightarrow \dfrac{{f}_{w}}{{f}_{a}}=\dfrac{\left(1.5-1\right)}{\left(\dfrac{1.5\times 3}{4}-1\right)}\Rightarrow {f}_{w}=\dfrac{0.5}{0.125}\times 18=72cm
Thus, change in focal length of the lens is 72−18=54cm72-18=54cm

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