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

(a) Define the power of a lens.

(b) What is the SI unit of a power?

(c) Find the position of the image formed by the lens combination given in the following figure.

The combination of lens

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 \[{{r}_{1}}+{{r}_{2}}=A,\,\,\text{here}\,\,{{r}_{2}}={{i}_{c}}\] \[\therefore \]      \[{{r}_{1}}+{{i}_{c}}=A\] \[\Rightarrow \]   \[{{r}_{1}}41{}^\circ =60{}^\circ \] \[\Rightarrow \] \[{{r}_{1}}=16{}^\circ \] Using the formula, \[\mu =\frac{sin\,\,{{i}_{c}}}{\sin \,\,{{r}_{1}}}\] Or         \[sin\,{{i}_{1}}=1.524\,\,sin\,19{}^\circ =1.524\times 0.3256\] Or         \[{{i}_{1}}=sin-(0.4962)\] \[\Rightarrow \] \[{{i}_{1}}=29{}^\circ 75\grave{\ }\] Thus, incident angle should be \[29{}^\circ 75\grave{\ }\] Or (a) Power of a lens Power of a 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. \[P=1/\,f\,(m)\] (b) SI unit of a power SI unit of a power is diopter (D) or \[{{m}^{-1}}.\] (c) Image formed by first lens, \[\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}=\frac{1}{{{f}_{1}}}\] \[\Rightarrow \] \[\frac{1}{{{v}_{1}}}-\frac{1}{(-\,30)}=\frac{1}{10}\] \[\Rightarrow \]   \[\frac{1}{{{v}_{1}}}+\frac{1}{30}=\frac{1}{10}\] \[\Rightarrow \] \[\frac{1}{{{v}_{1}}}=\frac{1}{10}-\frac{1}{30}\] \[\Rightarrow \]   \[{{v}_{1}}=15\,cm\] So, this is at a distance of \[(15-5)\,\,cm=10\,cm\]right of second lens. The image serves as virtual object for second lens, which means that the rays appear to come from it for the second lens, \[\frac{1}{{{v}_{2}}}-\frac{1}{10}=\frac{1}{-10}\] \[\Rightarrow \] \[{{v}_{2}}=\infty \] The virtual image is formed at infinite distance to the left of the second lens. This acts as an object for third lens. \[\frac{1}{{{v}_{3}}}-\frac{1}{{{u}_{3}}}=\frac{1}{{{f}_{3}}}\] \[\Rightarrow \] \[\frac{1}{{{v}_{3}}}=\frac{1}{\infty }+\frac{1}{30}\] \[\Rightarrow \]               \[{{v}_{3}}=30\,\,cm\] The final image is formed \[30\,cm\]to the right of the third lens.