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QJEEM22Phff06: JEE Main 2022 Physics - Class 12 JEEM22Phff06 Solution: The graph between and for a thin convex lens in order to…

The graph between 1u\frac{1}{u} and 1v\frac{1}{v} for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is 1.51.5 and its both the surfaces have same radius of curvature RR . The value of RR will be ____________ cm.\mathrm{cm} . (where u=u= object distance, v=v= image distance)

JEE Main 2022 (Online) 26th July Morning Shift Physics - Geometrical Optics Question 88 English

Step-by-Step Verified Solution

f=10f = 10 cm

1f=(μ−1)(1R−1−R){1 \over f} = (\mu - 1)\left( {{1 \over R} - {1 \over { - R}}} \right)

110=1.5−11×2R{1 \over {10}} = {{1.5 - 1} \over 1} \times {2 \over R}

110=1R{1 \over {10}} = {1 \over R}

R=10R = 10 cm

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