The graph between 1/u and 1/v for a thin convex lens in order to… | JEE Main 2022 (26 Jul, Shift 1)

Q67JEE Main 2022 · 26 JUL · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

The graph between $\frac{1}{u}$ and $\frac{1}{v}$ 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 $1.5$ and its both the surfaces have same radius of curvature $R$ . The value of $R$ will be ____________ $\text{cm.}$ (where $u =$ object distance, $v =$ image distance)

Numerical answer type. Work it out, then open the answer.

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Answer: 10

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ : at $\frac{1}{u} = 0$ , $\frac{1}{v} = \frac{1}{f} = 0.10\text{ cm}^{-1}$ , so $f = 10$ cm.
Equiconvex lens: $\frac{1}{f} = (\mu - 1)\frac{2}{R} = 0.5 \times \frac{2}{R} = \frac{1}{R}$
$R = f = 10$ cm

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