1. **Problem statement:** Consider a grating with 5 narrow slits separated by distance $d$, illuminated by monochromatic light of vacuum wavelength $\lambda$ inside glass of refractive index $n$. The source is at angle $\theta$ inside the glass, and the screen is far away at angle $\Psi$ in vacuum. We want to find the phase difference $\delta$ between adjacent slits and the intensity $I$ on the screen. 2. **Phase difference for $\theta=0$:** When $\theta=0$, the rays inside the glass are normal to the grating. The path difference between adjacent slits for rays observed at angle $\Psi$ in vacuum is $d \sin \Psi$. The phase difference is related to path difference by $$\delta = \frac{2\pi}{\lambda} \times \text{path difference} = \frac{2\pi d}{\lambda} \sin \Psi$$ 3. **Phase difference for arbitrary $\theta$:** When the source is at angle $\theta$ inside the glass, the incident wavefronts have a phase difference between slits of $$\delta_{inc} = \frac{2\pi}{\lambda/n} d \sin \theta = \frac{2\pi n d}{\lambda} \sin \theta$$ The outgoing phase difference at angle $\Psi$ in vacuum is $$\delta = \frac{2\pi d}{\lambda} \sin \Psi - \delta_{inc} = \frac{2\pi d}{\lambda} (\sin \Psi - n \sin \theta)$$ 4. **Intensity expression:** For $N=5$ slits, the intensity on the screen is given by the interference formula $$I = I_0 \left( \frac{\sin \left( \frac{N \delta}{2} \right)}{\sin \left( \frac{\delta}{2} \right)} \right)^2$$ where $I_0$ is the intensity from one slit. 5. **Sketch of intensity vs $\sin \Psi$:** The principal maxima occur when $\delta = 2\pi m$, i.e. $$\sin \Psi = n \sin \theta + m \frac{\lambda}{d}, \quad m=0, \pm 1, \pm 2, ...$$ Minima occur where numerator zeros but denominator does not, between maxima. 6. **Intensity with two outside slits blocked:** Now $N=3$ slits (slits 2,3,4). The intensity is $$I = I_0 \left( \frac{\sin \left( \frac{3 \delta}{2} \right)}{\sin \left( \frac{\delta}{2} \right)} \right)^2$$ 7. **Sketch comparison:** The 3-slit pattern has fewer and broader maxima than the 5-slit pattern. The maxima locations remain the same but the maxima intensity decreases roughly proportional to $N^2$ (so smaller for 3 slits). **Final answers:** - (a) $\delta = \frac{2\pi d}{\lambda} \sin \Psi$ - (b) $\delta = \frac{2\pi d}{\lambda} (\sin \Psi - n \sin \theta)$ - (c) $I = I_0 \left( \frac{\sin \left( \frac{N \delta}{2} \right)}{\sin \left( \frac{\delta}{2} \right)} \right)^2$ - (e) For 3 slits: $I = I_0 \left( \frac{\sin \left( \frac{3 \delta}{2} \right)}{\sin \left( \frac{\delta}{2} \right)} \right)^2$