(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? |
Answer:
(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 \[=|{{u}_{e}}|+|{{v}_{0}}|\] \[=4.17+84=10.57\,cm\]
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