The graph between (1/ u ) and (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 length is 1.5 and its both the surfaces have same radius of curvatures R. The value of R will be cm. (Where u = object distance, v = image distance)

Question

The graph between (1/ u ) and (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 length is 1.5 and its both the surfaces have same radius of curvatures R. The value of R will be cm. (Where u = object distance, v = image distance) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/69e24a77eca8be6e8f201e47736623b9-.png" /> For point B, (1/ u )=-0.10 cm -1, (1/ v )=0 ∴Thus, u =-10 cm , v =∞ i.e. f =10 cm ⇒ (1/10 cm )=(1.5-1)((2/ R ))=(1/ R ) ⇒ R =10 cm

For point B,u1​=−0.10cm−1,v1​=0 ∴Thus, u=−10cm,v=∞ i.e. f=10cm ⇒10cm1​=(1.5−1)(R2​)=R1​ ⇒R=10cm

For point $B, \frac{1}{u} = - 0.10 c m^{- 1}, \frac{1}{v} = 0$
$∴$ Thus, $u = - 10 c m, v = \infty$
i.e. $f = 10 c m$
$\Rightarrow \frac{1}{10 c m} = (1.5 - 1) (\frac{2}{R}) = \frac{1}{R}$
$\Rightarrow R = 10 c m$