NCERT Exemplar · Class 12 Physics Ray Optics and Optical Instruments

Let the two ends of the object be at distance \(\displaystyle u_1=u-L / 2\) and \(\displaystyle u_2=u+L / 2\), respectively, so that \(\displaystyle \left|u_1-u_2\right|=L\). Let the image of the two ends be formed at \(\displaystyle v_1\) and \(\displaystyle v_2\), so that the image length would be \(\displaystyle L^{\prime}=\left|v_1-v_2\right|\). Since \(\displaystyle \frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) or \(\displaystyle v=\frac{f u}{u-f}\) the image of the two ends will be at \(\displaystyle v_1=\frac{f(u-L / 2)}{u-f-L / 2}, v_2=\frac{f(u+L / 2)}{u-f+L / 2}\) Hence \[L^{\prime}=\left|v_1-v_2\right|=\frac{f^2 L}{(u-f)^2 \times L^2 / 4} \] Since the object is short and kept away from focus, we have \[L^2 / 4 \ll(u-f)^2 \] Hence finally \[L^{\prime}=\frac{f^2}{(u-f)^2} L . \]

If there was no cut then the object would have been at a height of $\displaystyle 0.5$ cm from the principal axis $\displaystyle 00$′. Consider the image for this case. \[\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \therefore \frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50} \\ & \therefore v=50 \mathrm{~cm} . \end{aligned} \] Magnification is \(\displaystyle \mathrm{m}=\frac{v}{u}=-\frac{50}{50}=-1\). Thus the image would have been formed at $\displaystyle 50$ cm from the pole and $\displaystyle 0.5$ cm below the principal axis. Hence with respect to the X axis passing through the edge of the cut lens, the co-ordinates of the image are \[(50 \mathrm{~cm},-1 \mathrm{~cm}) \]

From the reversibility of \(\displaystyle u\) and \(\displaystyle v\), as seen from the formula for lens, \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] It is clear that there are two positions for which there shall be an image on the screen. Let the first position be when the lens is at O. Given \(\displaystyle -u+v=D\) \[\Rightarrow u=-(D-v) \] Placing it in the lens formula \[\frac{1}{D-v}+\frac{1}{v}=\frac{1}{f} \] \[\begin{aligned} & \Rightarrow \frac{v+D-v}{(D-v) v}=\frac{1}{f} \\ & \Rightarrow v^2-D v+\mathrm{D} f=0 \\ & \Rightarrow v=\frac{D}{2} \pm \frac{\sqrt{\mathrm{D}^2-4 D f}}{2} \end{aligned} \] \[u=-(D-v)=-\left(\frac{D}{2} \pm \frac{\sqrt{D^2-4 D f}}{2}\right) \] Thus, if the object distance is \(\displaystyle \frac{D}{2}-\frac{\sqrt{D^2-4 D f}}{2}\) then the image is at \[\frac{D}{2}+\frac{\sqrt{D^2-4 D f}}{2} \] If the object distance is \(\displaystyle \frac{D}{2}+\frac{\sqrt{D^2-4 D f}}{2}\), then the image is at \[\frac{D}{2}-\frac{\sqrt{D^2-4 D f}}{2} . \] The distance between the poles for these two object distances is \[\frac{D}{2}+\frac{\sqrt{D^2-4 D f}}{2}-\left(\frac{D}{2}-\frac{\sqrt{D^2-4 \mathrm{D} f}}{2}\right)=\sqrt{D^2-4 D f} \] Let \(\displaystyle d=\sqrt{D^2-4 D f}\) If \(\displaystyle u=\frac{D}{2}+\frac{d}{2}\) then the image is at \(\displaystyle v=\frac{D}{2}-\frac{d}{2}\). ∴ The magnification \(\displaystyle m_1=\frac{\mathrm{D}-d}{D+d}\) If \(\displaystyle u=\frac{\mathrm{D}-d}{2}\) then \(\displaystyle v=\frac{\mathrm{D}+d}{2}\) ∴ The magnification \(\displaystyle m_2=\frac{\mathrm{D}+d}{\mathrm{D}-d}\) - Thus \(\displaystyle \frac{m_2}{m_1}=\left(\frac{\mathrm{D}+d}{\mathrm{D}-d}\right)^2\).

Let \(\displaystyle d\) be the diameter of the disc. The spot shall be invisible if the incident rays from the dot at O to the surface at \(\displaystyle \frac{d}{2}\) are at the critical angle. Let \(\displaystyle i\) be the angle of incidence. Then \(\displaystyle \sin i=\frac{1}{\mu}\) Now, \(\displaystyle \frac{d / 2}{h}=\tan i\) \[\begin{aligned} & \Rightarrow \frac{d}{2}=h \tan i=h\left[\sqrt{\mu^2-1}\right]^{-1} \\ & \therefore d=\frac{2 h}{\sqrt{\mu^2-1}} . \end{aligned} \]