(i) SI unit of power of lens: Dioptre is the SI unit of power of a lens. The power of the lens is the reciprocal of its focal length measured in metre(m). (ii) (iii) Refractive index $=\mu=1.5$ Radius of curvature of convex side $=25\text{ cm}$ $=\frac{25}{100}\text{ m}$ Radius of curvature of plane side $=\infty$ Applying lens maker's formula, $\frac{1}{f} = (\mu-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ Or, $\frac{1}{f}=(1.5-1)\left(\frac{1}{\infty}+\frac{100}{25}\right)$ Or, $\frac{1}{f} = 0.5 \times \left(\frac{100}{25}\right)$ $\therefore f = \frac{25}{50}\text{ m} = 50\text{ m}$ (iv) Focal Length $= f=50\text{ cm}$ Object distance $= u=-50\text{ cm}$ When the object distance $=$ focal length, then the image will be formed at infinity. Image will be highly magnified, real and inverted. OR (b) (i) Distance of 2nd bright fringe from central maximum $= \frac{2\lambda D}{d}$ $= \frac{2 \times 600 \times 10^{-9} \times 1}{0.6 \times 10^{-3}}$ (ii) If $n^{\text{th}}$ bright fringe due to $600\text{ nm}$ coincides with $(n+1)$ bright fringe due to $480\text{ nm}$, then $\frac{n\lambda_1 D}{d} = \frac{(n+1)\lambda_2 D}{d}$ Or $n\lambda_1 D = (n+1)\lambda_2 D$ Or, $\frac{n}{(n+1)} = \frac{\lambda_2}{\lambda_1}$ Or, $\frac{n}{n+1} = \frac{480}{600}$ $\therefore n=4$ So, the least distance from central maximum $= \frac{4 \times 600 \times 10^{-9} \times 1}{0.6 \times 10^{-3}} = 40 \times 10^{-4}\text{ m}$