A 20 D lens is used as a magnifier. Where should the object be placed to obtain maximum angular magnification? (Given, D=25 cm).

Sure, here are the step by step solutions: Given: Focal length of the lens, $f=\frac{1}{20}m=\frac{100}{20}cm=5cm$ To find: Maximum angular magnification and the object distance $u_0$ for which it occurs Formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the lens Let the distance of the object from the lens be $u_0$ and the distance of the final image from the lens be $D$ From the formula, we have: $\frac{1}{D}-\frac{1}{u_0}=\frac{1}{f}$$\frac{1}{D}=\frac{1}{f}+\frac{1}{u_0}$ Maximum angular magnification is obtained when final image is formed at $D$. Hence, we need to find the value of $u_0$ for which $D$ is maximum. To find the maximum value of $D$, we need to differentiate the equation $\frac{1}{D}=\frac{1}{f}+\frac{1}{u_0}$ with respect to $u_0$ and equate it to zero. $\frac{d}{d{u}_0}\left(\frac{1}{D}\right)=\frac{d}{d{u}_0}\left(\frac{1}{f}+\frac{1}{u_0}\right)$$-\frac{1}{D^2}\frac{dD}{d{u}_0}=-\frac{1}{u_0^2}$$\frac{dD}{d{u}_0}=\frac{D^2}{u_0^2}$ At maximum $D$, $\frac{dD}{d{u}_0}=0$ Therefore, $\frac{D^2}{u_0^2}=0$$\Rightarrow D=0$ Thus, the maximum angular magnification is obtained when the final image is formed at infinity. Substituting $D=\infty$ in the equation $\frac{1}{D}=\frac{1}{f}+\frac{1}{u_0}$, we get: $\frac{1}{\infty }=\frac{1}{5}+\frac{1}{u_0}$$\Rightarrow \frac{1}{u_0}=\frac{1}{\infty }-\frac{1}{5}$$\Rightarrow \frac{1}{u_0}=-\frac{1}{5}$$\Rightarrow u_0=-5cm$ Therefore, the object distance $u_0$ for which the maximum angular magnification is obtained is $-5cm$.