JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 25)
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 ____________ $$\mathrm{cm} .$$ (where $$u=$$ object distance, $$v=$$ image distance)
_26th_July_Morning_Shift_en_25_1.png)
Answer
10
Explanation
$$f = 10$$ cm
$${1 \over f} = (\mu - 1)\left( {{1 \over R} - {1 \over { - R}}} \right)$$
$${1 \over {10}} = {{1.5 - 1} \over 1} \times {2 \over R}$$
$$ {1 \over {10}} = {1 \over R}$$
$$R = 10$$ cm
Comments (0)
