• # question_answer (i) What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses. (ii) At what angle should a ray of light be incident on the face of a prism of refracting angle $60{}^\circ ,$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524. Or (i) Define the power of lens. (ii) An angular magnification (magnifying power) of 24 X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

(i) Given, focal length of convex lens, ${{f}_{1}}=30\,cm$ Focal length of concave lens, ${{f}_{2}}=-20\,cm$ Using the formula of combination of lenses $\frac{1}{f}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\Rightarrow f=-\,60\,cm$Since, the focal length of combination is negative in nature. So, the combination behaves like a diverging lens, i.e. as a concave lens. (ii) Angle of prism, $A=60{}^\circ$ Refractive index of prism, $\mu =1.524$ Let i be the angle of incidence. The angle of incidence at the other surface is equal to the critical angle ${{i}_{c}}$  because it just suffers total internal refraction. $\therefore$      $\sin {{i}_{c}}=\frac{1}{\mu }=\frac{1}{1.524}=0.6561$ $\Rightarrow$   ${{i}_{c}}=41{}^\circ$ For a prism ${{r}_{1}}+{{i}_{c}}=A\Rightarrow {{r}_{1}}+41{}^\circ =60{}^\circ$ $\Rightarrow$   ${{r}_{1}}=19{}^\circ$ Using the formula, $\mu =\frac{\sin {{i}_{1}}}{\sin {{r}_{1}}}$ or         $\sin {{i}_{1}}=1.524\sin 19{}^\circ =1.524\times 0.3256$ or         ${{i}_{1}}={{\sin }^{-1}}(0.4962)$             ${{i}_{1}}=29{}^\circ 75'$ Thus, incident angle should be $29{}^\circ 75'.$ Or (i) Power of lens Power of lens is the ability to diverge or converge the light rays incident on it. It is defined as the reciprocal of focal length in metres.             $P=\frac{1}{f(m)}$ It's SI unit of power is diopter (D) or ${{m}^{-1}}.$ (ii) We assume the microscope in common usage, i.e. the final image is formed at the least distance of distinct vision, D=25 cm, ${{f}_{e}}=5\,cm$ $\therefore$ Angular magnification of the eyepiece is             ${{m}_{e}}=1+\frac{D}{{{f}_{e}}}=1+\frac{25}{5}=6$ As total magnification, $m={{m}_{e}}\times {{m}_{o}}$ $\therefore$ Angular magnification of the objective is              ${{m}_{o}}=\frac{m}{{{m}_{e}}}=\frac{24}{6}=4$ As real image is formed by the objective, therefore ${{m}_{o}}=\frac{{{v}_{o}}}{{{u}_{o}}}=-4$ or ${{v}_{0}}=-\,4{{u}_{o}},$ ${{f}_{o}}=125\,cm$ Now      $\frac{1}{{{v}_{o}}}-\frac{1}{{{u}_{o}}}=\frac{1}{{{f}_{o}}}$ or         $\frac{1}{-\,4{{u}_{o}}}-\frac{1}{{{u}_{o}}}=\frac{1}{125}$ or         $\frac{-5}{4{{u}_{o}}}=\frac{1}{125}$ or         ${{u}_{o}}=\frac{5\times 125}{4}=-1.6\,cm$ Thus, the object should be held at 1.5 cm in front of the objective lens. Also,     ${{v}_{o}}=-4{{u}_{o}}=-4\times (-1.6)=6.4\,cm$ As,        $\frac{1}{{{V}_{e}}}-\frac{1}{{{u}_{e}}}=\frac{1}{{{f}_{e}}}$ $\therefore$      $\frac{1}{{{u}_{e}}}=\frac{1}{{{v}_{e}}}-\frac{1}{{{f}_{e}}}=\frac{1}{-25}-\frac{1}{5}$                                     $[\therefore \,{{v}_{e}}=-D=-25\,cm]$             $=\frac{-1-5}{25}=-\frac{6}{25}$ or         ${{u}_{e}}=\frac{-25}{6}=-\,4.17cm$ $\therefore$ Separation between the objective and the eyepiece                         $=|{{u}_{e}}|+|{{v}_{0}}|$ $=4.17+84=10.57\,cm$