Let's derive the intensity pattern for a **three-slit grating** as requested. ### **Step 1: Setup** - **Number of slits, \( N = 3 \)** - **Let each slit have the same amplitude \( E_0 \).** - **Phase difference between adjacent slits:** \[ \phi = \frac{2\pi d \sin \theta}{\lambda} \] where \( d \) = slit separation, \( \theta \) = angle, \( \lambda \) = wavelength. ### **Step 2: Write the resultant electric field** For three slits, the field at a point on the screen is the sum of the fields from each slit, each with a phase difference \( \phi \): \[ E = E_0 \left[ 1 + e^{i\phi} + e^{i2\phi} \right] \] ### **Step 3: Simplify the sum** This is a geometric series: \[ E = E_0 \sum_{n=0}^{2} e^{in\phi} = E_0 \frac{1 - e^{i3\phi}}{1 - e^{i\phi}} \] But let's expand \( 1 + e^{i\phi} + e^{i2\phi} \) directly: \[ E = E_0 \left[ 1 + e^{i\phi} + e^{i2\phi} \right] \] ### **Step 4: Use Euler's formula to expand** Let’s use the sum formula for exponentials: \[ E = E_0 [1 + e^{i\phi} + e^{i2\phi}] \] \[ = E_0 \left[ 1 + (\cos\phi + i\sin\phi) + (\cos 2\phi + i\sin 2\phi) \right] \] \[ = E_0 [1 + \cos\phi + \cos 2\phi + i(\sin\phi + \sin 2\phi)] \] ### **Step 5: Intensity is proportional to the square of the amplitude** \[ I \propto |E|^2 \] Let’s compute \( |E|^2 \): \[ |E|^2 = \left[1 + \cos\phi + \cos 2\phi \right]^2 + \left[\sin\phi + \sin 2\phi \right]^2 \] ### **Step 6: Expand the squares** Let’s use trigonometric identities to expand further. #### **Sum of cosines:** \[ 1 + \cos\phi + \cos 2\phi \] Recall: \[ \cos 2\phi = 2\cos^2\phi - 1 \] So, \[ 1 + \cos\phi + \cos 2\phi = 1 + \cos\phi + (2\cos^2\phi - 1) = \cos\phi + 2\cos^2\phi \] #### **Sum of sines:** \[ \sin\phi + \sin 2\phi \] Recall: \[ \sin 2\phi = 2\sin\phi\cos\phi \] So, \[ \sin\phi + \sin 2\phi = \sin\phi + 2\sin\phi\cos\phi = \sin\phi(1 + 2\cos\phi) \] ### **Step 7: Write out \( |E|^2 \)** \[ |E|^2 = [\cos\phi + 2\cos^2\phi]^2 + [\sin\phi(1 + 2\cos\phi)]^2 \] Expand both terms: #### **First term:** \[ (\cos\phi + 2\cos^2\phi)^2 = \cos^2\phi + 4\cos^3\phi + 4\cos^4\phi \] But let's keep it grouped for now. #### **Second term:** \[ [\sin\phi(1 + 2\cos\phi)]^2 = \sin^2\phi(1 + 4\cos\phi + 4\cos^2\phi) \] ### **Step 8: Add and simplify** Combine terms: \[ |E|^2 = (\cos\phi + 2\cos^2\phi)^2 + \sin^2\phi (1 + 4\cos\phi + 4\cos^2\phi) \] Expand the squares: 1. \((\cos\phi + 2\cos^2\phi)^2 = \cos^2\phi + 4\cos^3\phi + 4\cos^4\phi\) 2. \(\sin^2\phi (1 + 4\cos\phi + 4\cos^2\phi) = \sin^2\phi + 4\sin^2\phi\cos\phi + 4\sin^2\phi\cos^2\phi\) Sum all terms: \[ |E|^2 = \cos^2\phi + 4\cos^3\phi + 4\cos^4\phi + \sin^2\phi + 4\sin^2\phi\cos\phi + 4\sin^2\phi\cos^2\phi \] But this looks complicated. Let's use the original sum formula for \( N \) slits: \[ E = E_0 \frac{\sin (N\phi/2)}{\sin (\phi/2)} \] For \( N = 3 \): \[ E = E_0 \frac{\sin (3\phi/2)}{\sin (\phi/2)} \] So, \[ I = I_0 \left[ \frac{\sin (3\phi/2)}{\sin (\phi/2)} \right]^2 \] But the question is asking for the form: \[ I = \frac{1}{9} I_m (1 + 4\cos\phi + 4\cos^2\phi) \] Let's try to rewrite the above in this form. ### **Step 9: Use trigonometric identity for \(\sin(3x)\)** \[ \sin(3x) = 3\sin x - 4\sin^3 x \] Let \( x = \phi/2 \): \[ \sin(3\phi/2) = 3\sin(\phi/2) - 4\sin^3(\phi/2) \] So, \[ \left[ \frac{\sin(3\phi/2)}{\sin(\phi/2)} \right]^2 = \left[ 3 - 4\sin^2(\phi/2) \right]^2 \] \[ = 9 - 24\sin^2(\phi/2) + 16\sin^4(\phi/2) \] Now, use the identity: \[ \sin^2(\phi/2) = \frac{1 - \cos\phi}{2} \] So, \[ \sin^4(\phi/2) = \left( \frac{1 - \cos\phi}{2} \right)^2 = \frac{1 - 2\cos\phi + \cos^2\phi}{4} \] Plug these into the equation: \[ 9 - 24\left(\frac{1 - \cos\phi}{2}\right) + 16\left(\frac{1 - 2\cos\phi + \cos^2\phi}{4}\right) \] \[ = 9 - 12(1 - \cos\phi) + 4(1 - 2\cos\phi + \cos^2\phi) \] \[ = 9 - 12 + 12\cos\phi + 4 - 8\cos\phi + 4\cos^2\phi \] \[ = (9 - 12 + 4) + (12\cos\phi - 8\cos\phi) + 4\cos^2\phi \] \[ = 1 + 4\cos\phi + 4\cos^2\phi \] ### **Step 10: Normalize** For a maximum intensity \( I_m \), the central maximum (where \( \phi = 0 \)), this expression becomes: \[ I = I_m \frac{1}{9} (1 + 4\cos\phi + 4\cos^2\phi) \] ### **Final Answer:** \[ \boxed{ I = \frac{1}{9} I_m (1 + 4\cos\phi + 4\cos^2\phi) } \] where \( \phi = \frac{2\pi d \sin\theta}{\lambda} \), and \( a \ll \lambda \). --- **Summary of steps:** 1. Write the field sum for three slits. 2. Use the geometric sum or trigonometric identities to expand. 3. Use the triple-angle formula for sine. 4. Simplify using trigonometric identities. 5. Normalize to get the final expression for intensity.

Let's derive the expression for the intensity pattern of a three-slit grating step-by-step, incorporating detailed calculations and the given data. --- ### **Given Data:** - Number of slits: \( N = 3 \) - Slit separation: \( d \) - Wavelength: \( \lambda \) - The phase difference: \( \phi = \frac{2\pi d \sin \theta}{\lambda} \) - \( a \ll \lambda \), which indicates slit width is small, and the primary pattern depends on \( d \) and \( \lambda \). --- ### **Objective:** Derive the intensity pattern: \[ I = \frac{1}{9} I_m (1 + 4 \cos \phi + 4 \cos^2 \phi) \] --- ### **Step 1: Electric field from three slits** Each slit contributes an electric field \( E_0 \) at the observation point, with a phase shift: \[ E_1 = E_0 \] \[ E_2 = E_0 e^{i \phi} \] \[ E_3 = E_0 e^{i 2\phi} \] Total electric field: \[ E_\text{total} = E_0 \left( 1 + e^{i\phi} + e^{i 2\phi} \right) \] --- ### **Step 2: Sum of the fields** Recognize this as a geometric series: \[ E_\text{total} = E_0 \frac{1 - e^{i 3 \phi}}{1 - e^{i \phi}} \] The magnitude squared (which gives intensity) is: \[ I \propto |E_\text{total}|^2 = E_0^2 \left| \frac{1 - e^{i 3 \phi}}{1 - e^{i \phi}} \right|^2 \] --- ### **Step 3: Simplify numerator and denominator** Using the identity: \[ |1 - e^{i \alpha}| = 2 |\sin (\alpha/2)| \] we get: \[ |1 - e^{i 3 \phi}| = 2 |\sin \frac{3\phi}{2}| \] \[ |1 - e^{i \phi}| = 2 |\sin \frac{\phi}{2}| \] Therefore, \[ I \propto E_0^2 \left( \frac{2 |\sin \frac{3\phi}{2}|}{2 |\sin \frac{\phi}{2}|} \right)^2 = E_0^2 \left( \frac{\sin \frac{3\phi}{2}}{\sin \frac{\phi}{2}} \right)^2 \] --- ### **Step 4: Express intensity** Let \( I_0 \) be the maximum intensity (when all fields are in phase). Then: \[ I = I_0 \left( \frac{\sin \frac{3\phi}{2}}{\sin \frac{\phi}{2}} \right)^2 \] --- ### **Step 5: Expand numerator using the sine triple-angle identity** Recall: \[ \sin 3x = 3 \sin x - 4 \sin^3 x \] Set \( x = \frac{\phi}{2} \): \[ \sin \frac{3\phi}{2} = 3 \sin \frac{\phi}{2} - 4 \sin^3 \frac{\phi}{2} \] Thus, \[ \frac{\sin \frac{3\phi}{2}}{\sin \frac{\phi}{2}} = 3 - 4 \sin^2 \frac{\phi}{2} \] Squaring both sides: \[ \left( \frac{\sin \frac{3\phi}{2}}{\sin \frac{\phi}{2}} \right)^2 = \left( 3 - 4 \sin^2 \frac{\phi}{2} \right)^2 \] --- ### **Step 6: Express in terms of \( \cos \phi \)** Use the identity: \[ \sin^2 \frac{\phi}{2} = \frac{1 - \cos \phi}{2} \] Substitute into the expression: \[ \left( 3 - 4 \times \frac{1 - \cos \phi}{2} \right)^2 = \left( 3 - 2 (1 - \cos \phi) \right)^2 \] Simplify: \[ = \left( 3 - 2 + 2 \cos \phi \right)^2 = (1 + 2 \cos \phi)^2 \] Expanding: \[ (1 + 2 \cos \phi)^2 = 1 + 4 \cos \phi + 4 \cos^2 \phi \] --- ### **Step 7: Final expression for intensity** Thus, the intensity pattern is: \[ I = I_0 \times \left( 1 + 4 \cos \phi + 4 \cos^2 \phi \right) \] The maximum intensity occurs at \( \phi = 0 \), giving: \[ I_m = I_0 \times (1 + 4 + 4) = I_0 \times 9 \] Therefore, normalize to get the pattern: \[ I = \frac{I_m}{9} \left( 1 + 4 \cos \phi + 4 \cos^2 \phi \right) \] --- ### **### **Final answer:** \[ \boxed{ \boxed{ I = \frac{1}{9} I_m \left( 1 + 4 \cos \phi + 4 \cos^2 \phi \right) } } \] where \[ \phi = \frac{2 \pi d \sin \theta}{\lambda} \] and \( a \ll \lambda \). --- ### **Summary:** - The derivation uses the sum of fields from three slits. - The key step involves the sine triple-angle identity. - The normalized intensity pattern reveals the interference fringes characteristic of a three-slit grating. If you have specific values for \( d \), \( \lambda \), or \( \theta \), you can substitute into the expression for \( \phi \) to compute the intensity at different angles.