(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 $=\mathrm{mu}=1.5$
Radius of curvature of convex side $=25\mathit{c}\mathit{m}$
$=\frac{25}{100}\mathit{m}$
Radius of curvature of plane side $=\mathit{\infty }$
Applying lens maker's formula,
$\frac{1}{\mathit{f}}=\left(\mathit{\mu }-1\right)\left(\frac{1}{{\mathit{R}}_1}-\frac{1}{{\mathit{R}}_2}\right)$
Or, $\frac{1}{\mathit{f}}=\left(1.5-1\right)\left(\frac{1}{\mathit{\infty }}+\frac{100}{25}\right)$
Or, $\frac{1}{\mathit{f}}=0.5\times \left(\frac{100}{25}\right)$
$\therefore \mathit{f}=\frac{25}{50}\mathit{m}=50\mathit{m}$
(iv) Focal Length $=\mathit{f}=50\mathit{c}\mathit{m}$
Object distance $=\mathit{u}=-50\mathit{c}\mathit{m}$
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 $=\frac{2\mathit{\lambda }\mathit{D}}{\mathit{d}}$
$=\frac{2\times 600\times {10}^{-9}\times 1}{0.6\times {10}^{-3}}$
(ii) If ${\mathit{n}}^{\mathit{t}\mathit{h}}$ bright fringe due to $600\mathit{n}\mathit{m}$ coincides with $\left(\mathit{n}+1\right)$ bright fringe due to $480\mathit{n}\mathit{m}$, then
$\frac{\mathit{n}{\mathit{\lambda }}_1\mathit{D}}{\mathit{d}}=\frac{\left(\mathit{n}+1\right){\mathit{\lambda }}_2\mathit{D}}{\mathit{d}}$
Or $\mathit{n}{\mathit{\lambda }}_1\mathit{D}=\left(\mathit{n}+1\right){\mathit{\lambda }}_2\mathit{D}$
Or, $\frac{\mathit{n}}{\left(\mathit{n}+1\right)}=\frac{{\mathit{\lambda }}_2}{{\mathit{\lambda }}_1}$
Or, $\frac{\mathit{n}}{\left(\mathit{n}+1\right)}=\frac{480}{600}$
$\therefore \mathit{n}=4$
So, the least distance from central maximum
$=\frac{4\times 600\times {10}^{-9}\times 1}{0.6\times {10}^{-3}}=40\times {10}^{-4}\mathit{m}$