(a) Two characteristic features of distinction Derivation of the expression for the intensity (b) Calculation of separation between the first order (a) (Any two of the following) (i) Interference pattern has number of equally spaced bright and dark bands while diffraction pattern has central bright maximum which is twice as wide as the other maxima. (ii) Interference is obtained by the superposing two waves originating from two narrow slits. The diffraction pattern is the superposition of the continuous family of waves originating from each point on a single slit. (iii) In interference pattern, the intensity of all bright fringes is same, while in diffraction pattern intensity of bright fringes go on decreasing with the increasing order of the maxima. (iv) In interference pattern, the first maximum falls at an angle of $\;\frac{\lambda}{a}$. where $a$ is the separation between two narrow slits, while in diffraction pattern, at the same angle first minimum occurs. (where ' $a$ ' is the width of single slit.) Displacement produced by source $S_1$ $Y_1 = a \cos \omega t$ Displacement produced by the other source ' $S_2^{'}$ $\gamma_2 = a \cos (\omega t + \phi)$ Resultant displacement $Y = Y_1 + Y_2$ $\; = a [ \cos \omega t + \cos (\omega t + \phi)$ $\; = 2 a \cos \left( \frac{\phi}{2} \right) \cos \left( \omega t + \frac{\phi}{2} \right)$ Amplitude of resultant wave $A = 2 a \cos \left( \frac{\phi}{2} \right)$ Intensity $\;\; I \propto A^2$ $I = K A^2 = K 4 a^2 \cos^2 \left( \frac{\phi}{2} \right)$ (b) Distance of first order minima from centre of the central maxima $= X_{D_1} = \frac{\lambda \cdot D}{a}$ Distance of third order maxima from centre of the central maxima $X_{B_3} = \frac{7 D \lambda}{2 a}$ $\therefore$ Distance between first order minima and third order maxima $= X_{B_3} - X_{D_1}$ $= \frac{7 D \lambda}{2 a} - \frac{\lambda D}{a}$ $= \frac{5 D \lambda}{2 a}$ $= \frac{5 \times 620 \times 10^{-9} \times 1.5}{2 \times 3 \times 10^{-3}}$ $= 775 \times 10^{-6} \,\text{m}$ $= 7.75 \times 10^{-4} \,\text{m}$ OR (a) Two conditions of total internal reflection (b) Obtaining the relation (c) Calculating of the position of the final image (a) (i) Light travels from denser to rarer medium. (ii) Angle of incidence is more than the critical angle (b) For the Grazing incidence $\mu \sin i_C = \sin 90^{\circ}$ $\mu = \frac{1}{\sin i_C}$ (c) For convex lens of focal length $10\,\text{cm}$ $\;\;\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}$ $\;\;\frac{1}{10} = \frac{1}{v_1} - \frac{1}{-30} \Rightarrow v_1 = 15\,\text{cm}$ Object distance for concave lens $u_2 = 15 - 5 = 10\,\text{cm}$ $\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}$ $\frac{1}{-10} = \frac{1}{v_2} - \frac{1}{10}$ $v_2 = \infty$For third lens$\;\;\frac{1}{f_3} = \frac{1}{v_3} - \frac{1}{u_3}$$\;\;\frac{1}{30} = \frac{1}{v_3} - \frac{1}{\infty}$$\; v_3 = 30\,\text{cm}$