Q.In Young's double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is $K$ units. What is the intensity of light at a point where path difference is $\lambda/3$?

Showing the 12 most recent of 41 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.If the intensity of the central maximum in the Young's double slit experiment is $\mathrm{I}_0$, what will be the intensity at the same region when one of the slits is blocked by an opaque object? (A) $\frac{I_0}{2}$ (B) $I_0$ (C) $\frac{I_0}{4}$ (D) $\frac{I_0}{8}$

   ›Reveal solutionSolution

   In Young’s double-slit experiment, the central maximum intensity $I_0$ comes from the sum of two equal amplitudes. Blocking one slit removes one source, so the intensity drops to one-quarter of $I_0$. The correct option is (C).

   Concept & Intuition

   The key idea is that intensity is proportional to the square of the resultant amplitude. With both slits open, the waves from the two slits arrive in phase at the central point, so their amplitudes add constructively. If each slit alone would produce an amplitude $a$, the combined amplitude is $2a$, giving intensity $I_0 \propto (2a)^2 = 4a^2$. When one slit is blocked, only one wave of amplitude $a$ reaches the screen, so the intensity becomes $I \propto a^2$. Comparing: $I = I_0/4$.

   Step-by-step reasoning

   1. Define the amplitude from a single slit

      Let the amplitude of the wave from each slit (when alone) be $a$. The intensity from a single slit is $I_{\text{single}} \propto a^2$.

   2. Both slits open at the central maximum

      At the central point, the path difference is zero, so the waves are in phase. The resultant amplitude is $a + a = 2a$.

      Intensity is proportional to the square of amplitude:

   $I_0 \propto (2a)^2 = 4a^2.$

   3. One slit blocked Only one wave reaches the screen, so the amplitude is $a$. The intensity is

   $I \propto a^2.$

   4. Find the ratio From step 2: $I_0 = k \cdot 4a^2$ (where $k$ is the proportionality constant). …

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2. COMEDK 2026Set 2026-A1 markMCQ
   Q.In a single slit diffraction experiment, the diffraction pattern is observed on a screen placed at a distance of 2 m from the slit of 1 mm width. If the distance between the first dark fringe on either side of the central to right fringe is 2.2 mm . what is the wave length of the monochromatic light from a distance source, used in this experiment? (A) $3900^{\circ} \mathrm{A}$ (B) $11000^{\circ} \mathrm{A}$ (C) $1100^{\circ} \mathrm{A}$ (D) $5500^{\circ} \mathrm{A}$

   ›Reveal solutionSolution

   In single‑slit diffraction, the first minima occur at an angle satisfying $a \sin\theta = \lambda$. Using the small‑angle approximation and the given geometry, the wavelength is found to be $5500\,\text{Å}$, which corresponds to option (D).

   The key concept is single‑slit diffraction: when monochromatic light passes through a narrow slit, it spreads out and produces a pattern of alternating bright and dark fringes. The first dark fringe (minimum) on either side of the central maximum occurs when the path difference between light from the two edges of the slit equals exactly one wavelength. This condition is $a \sin\theta = \lambda$, where $a$ is the slit width and $\theta$ is the angular position of the first minimum.

   Why does this work? Because the problem gives the distance between the first dark fringes on either side — that is, the total width of the central bright region measured between the two first minima. On the screen, this distance is $2y$, where $y$ is the distance from the center to the first minimum. Using the small‑angle approximation ($\sin\theta \approx \tan\theta = y/D$), we can relate the geometry to the wavelength.

   1. Identify given quantities

      • Slit width: $a = 1\ \text{mm} = 1 \times 10^{-3}\ \text{m}$
      • Screen distance: $D = 2\ \text{m}$
      • Distance between first dark fringes on either side: $2y = 2.2\ \text{mm} = 2.2 \times 10^{-3}\ \text{m}$ Hence, the distance from the center to the first minimum is $y = 1.1 \times 10^{-3}\ \text{m}$.

   2. Apply the single‑slit minimum condition

      For the first minimum: $a \sin\theta = \lambda$.

      For small angles (since $y \ll D$), $\sin\theta \approx \tan\theta = \frac{y}{D}$.

      Therefore:

      $$$$

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3. COMEDK 2026Set 2026-A1 markMCQ
   Q.The width of the fringes obtained with a light of wave length $6.2 \times 10^{-8} \mathrm{~m}$ is 1.82 mm . If the whole apparatus is immersed in a liquid of refractive index 1.3 , what will be the width of the resulting fringe? (A) 1.4 mm (B) 0.71 mm (C) 2.8 mm (D) 1.82 mm

   ›Reveal solutionSolution

   The fringe width in Young’s double-slit experiment is proportional to the wavelength. Immersing the apparatus in a liquid reduces the wavelength by a factor equal to the refractive index, so the new fringe width is the original divided by 1.3, giving 1.4 mm.

   Concept & Intuition

   In Young’s double-slit experiment, the fringe width (the distance between consecutive bright or dark fringes) is given by

   $\beta = \frac{\lambda D}{d}$

   where $\lambda$ is the wavelength of light in the medium, $D$ is the distance from the slits to the screen, and $d$ is the slit separation. When the entire apparatus is immersed in a liquid, the only quantity that changes is the wavelength: the light’s frequency stays the same, but its speed decreases, so the wavelength in the liquid becomes $\lambda' = \frac{\lambda}{n}$, where $n$ is the refractive index of the liquid. Since $D$ and $d$ are unchanged, the new fringe width is simply $\beta' = \frac{\beta}{n}$.

   Step-by-step reasoning

   1. Recall the formula for fringe width In air (or vacuum), the fringe width is

   $\beta = \frac{\lambda D}{d}$

   Here $\lambda = 6.2 \times 10^{-8}\,\text{m}$ and $\beta = 1.82\,\text{mm}$.

   2. Identify what changes in the liquid The refractive index $n = 1.3$ tells us how much the speed of light reduces. The frequency $f$ remains constant, so the wavelength in the liquid is

   $\lambda' = \frac{\lambda}{n}$

   The slit separation $d$ and screen distance $D$ are physical dimensions that do not change when the apparatus is immersed.

   3. Write the new fringe width …

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4. COMEDK 2026Set 2026-M1 markMCQ
   Q.A diffraction pattern due to a single slit of width 0.12 mm is obtained with a blue green light of wavelength 500 nm . The angular separation between central maximum and second order secondary maximum of the diffraction pattern is (A) $0.042 \times 10^{-2} \mathrm{rad}$ (B) $1.042 \times 10^{-2} \mathrm{rad}$ (C) $1.042 \times 10^{-4} \mathrm{rad}$ (D) $0.042 \times 10^{-3} \mathrm{rad}$

   ›Reveal solutionSolution

   The angular separation between the central maximum and the second-order secondary maximum in single-slit diffraction is found using the condition for minima, not maxima. The second secondary maximum lies approximately halfway between the second and third minima, giving an angle of about $1.042 \times 10^{-2}$ rad. The correct option is (B).

   The key concept here is single-slit diffraction. Unlike the double-slit pattern, where bright fringes (maxima) occur at integer multiples of $\lambda/d$, in single-slit diffraction the bright fringes (secondary maxima) are not exactly halfway between minima. Instead, they are located at positions given approximately by $\sin\theta \approx (m + \frac12)\frac{\lambda}{a}$ for $m = 1, 2, 3, \dots$, but the exact positions require solving a transcendental equation. However, for most introductory problems, the approximation is sufficient, and the angular position of the $m$th secondary maximum is taken as midway between the $m$th and $(m+1)$th minima.

   Let’s work through it step by step.

   1. Identify the given data

      Slit width: $a = 0.12\ \text{mm} = 0.12 \times 10^{-3}\ \text{m} = 1.2 \times 10^{-4}\ \text{m}$

      Wavelength: $\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5 \times 10^{-7}\ \text{m}$

   2. Recall the condition for minima in single-slit diffraction

      For a slit of width $a$, dark fringes (minima) occur at angles $\theta$ satisfying:

   $a \sin\theta = n\lambda \quad (n = \pm 1, \pm 2, \pm 3, \dots)$

   Here $n$ is the order of the minimum. The central maximum corresponds to $n=0$.

   3. Locate the second secondary maximum

      The secondary maxima lie between the minima. The first secondary maximum is between the first and second minima ($n=1$ and $n=2$).

      The second secondary maximum lies between the second and third minima ($n=2$ and $n=3$).

      So we approximate its angular position as the average of the angles for $n=2$ and $n=3$ minima.

   4. Calculate the angles for the $n=2$ and $n=3$ minima

      For small angles, $\sin\theta \approx \theta$ (in radians).

      • For $n=2$:

   $\theta_2 \approx \frac{2\lambda}{a} = \frac{2 \times 5 \times 10^{-7}}{1.2 \times 10^{-4}} = \frac{10 \times 10^{-7}}{1.2 \times 10^{-4}} = \frac{10^{-6}}{1.2 \times 10^{-4}} = \frac{1}{1.2} \times 10^{-2} \approx 0.8333 \times 10^{-2}\ \text{rad}$

   • For $n=3$: …

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5. COMEDK 2026Set 2026-M1 markMCQ
   Q.In a Young's double slit experiment, the slit separation is 1.5 mm . The setup is illuminated simultaneously by light of wavelengths $6000 \mathop {\rm{A}}\limits^{\rm{o}}$ and $8000 \mathop {\rm{A}}\limits^{\rm{o}}$. The screen is placed at a distance of 1.5 m from the slits. It is observed that at a certain point $P$ on the screen which is 4.8 mm from the central maximum, fringes due to both the wavelengths coincide. Which of the following options are correct? A. Light of wavelength $6000\mathop {\rm{A}}\limits^{\rm{o}}$ produces a dark fringe and light of wavelength $8000 \mathop {\rm{A}}\limits^{\rm{o}}$ produces a bright fringe at P B. Light of wavelength $6000 \mathop {\rm{A}}\limits^{\rm{o}}$ produces a bright fringe and light of wavelength $8000 \mathop {\rm{A}}\limits^{\rm{o}}$ produces a dark fringe at P C. Light of wavelength $6000\mathop {\rm{A}}\limits^{\rm{o}}$ and light of wavelength $8000 \mathop {\rm{A}}\limits^{\rm{o}}$ both produce a dark fringe at P D. Light of wavelength $6000 \mathop {\rm{A}}\limits^{\rm{o}}$ and light of wavelength $8000\mathop {\rm{A}}\limits^{\rm{o}}$ both produce a bright fringe at P (A) B (B) D (C) A (D) C

   ›Reveal solutionSolution

   The path difference at $P$ is $4800$ nm, an integer multiple of both wavelengths ($8\lambda_1$ and $6\lambda_2$), so both produce a bright fringe at $P$ — statement D, which is option (B).

   Concept

   In Young's double-slit experiment the path difference at a distance $y$ from the central maximum is $\Delta=\dfrac{yd}{D}$. A wavelength gives a bright fringe where $\Delta=n\lambda$ (integer $n$) and a dark fringe where $\Delta=(n+\tfrac12)\lambda$.

   Solution

   1. Path difference. With $y=4.8$ mm, $d=1.5$ mm, $D=1.5$ m, $\Delta=\frac{yd}{D}=\frac{(4.8\times 10^{-3})(1.5\times 10^{-3})}{1.5}=4.8\times 10^{-6}\ \text{m}=4800\ \text{nm}.$ …

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6. COMEDK 2026Set 2026-M1 markMCQ
   Q.In a Young's double slit experiment, the slits are separated by 0.5 mm . Fringes are obtained on a screen which is placed at distance 1 m away from the slits. When the screen is moved 7 cm farther away, the fringe width changes by $63 \mu \mathrm{~m}$. The wavelength of light used in the experiment will be: (A) $4.5 \times 10^{-6} \mathrm{~m}$ (B) $4.5 \times 10^{-7} \mathrm{~m}$ (C) $0.5 \times 10^{-9} \mathrm{~m}$ (D) $0.5 \times 10^{-7} \mathrm{~m}$

   ›Reveal solutionSolution

   The fringe width in Young’s double slit is $\beta = \lambda D/d$. The change in fringe width when $D$ increases by $\Delta D$ gives $\Delta\beta = \lambda\,\Delta D / d$. Solving yields $\lambda = 4.5 \times 10^{-7}\,\mathrm{m}$, so the correct option is (B).

   The key idea is that fringe width $\beta$ is proportional to the screen distance $D$. When you move the screen farther, the fringes spread out linearly. The problem gives you the change in fringe width for a known change in distance, so you can directly solve for the wavelength without needing the original fringe width.

   ---

   1. Recall the formula for fringe width In Young’s double slit experiment, the fringe width (distance between consecutive bright or dark fringes) is

   $\beta = \frac{\lambda D}{d}$

   where $\lambda$ is the wavelength, $D$ is the distance from slits to screen, and $d$ is the slit separation.

   2. Identify the given quantities

      • Slit separation: $d = 0.5\ \mathrm{mm} = 0.5 \times 10^{-3}\ \mathrm{m}$
      • Initial screen distance: $D_1 = 1\ \mathrm{m}$
      • Screen moved farther by: $\Delta D = 7\ \mathrm{cm} = 0.07\ \mathrm{m}$
      • Change in fringe width: $\Delta\beta = 63\ \mu\mathrm{m} = 63 \times 10^{-6}\ \mathrm{m}$

   3. Express the change in fringe width

      The fringe width at the initial distance is $\beta_1 = \lambda D_1 / d$.

      At the new distance $D_2 = D_1 + \Delta D$, the fringe width is $\beta_2 = \lambda (D_1 + \Delta D)/d$.

      The change is

   $\Delta\beta = \beta_2 - \beta_1 = \frac{\lambda}{d} \Delta D$

   Notice that the initial distance $D_1$ cancels out — only the change in distance matters.

   4. Solve for the wavelength Rearranging:

   $\lambda = \frac{d \cdot \Delta\beta}{\Delta D}$

   Substitute the values:

   $\lambda = \frac{(0.5 \times 10^{-3})\,(63 \times 10^{-6})}{0.07}$ …

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7. KCET 2026Set C21 markMCQ
   Q.In Young's double slit experiment, how many maxima can be seen on a screen (including central maxima) if d = $\frac{5\lambda}{2}$ (where $\lambda$ is wavelength of light and d is distance between the two slits). (A) 5 (B) 4 (C) 7 (D) 1

   ›Reveal solutionSolution

   In Young's double slit experiment, bright fringes occur at path differences that are integer multiples of the wavelength; the number of orders possible is capped by the physical limit $\sin\theta \le 1$.

   Step 1 — Write the condition for a maximum

   A bright fringe (maximum) occurs where the path difference is an integer multiple of the wavelength:

   $d\sin\theta = n\lambda,\qquad n = 0,\pm1,\pm2,\dots$

   Step 2 — Apply the physical limit on $\sin\theta$

   Since $\sin\theta$ cannot exceed 1, the allowed orders must satisfy

   $|n| \le \frac{d}{\lambda} = \frac{5\lambda/2}{\lambda} = \frac{5}{2} = 2.5$ …

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8. COMEDK 2025Set 2025-A1 markMCQ
   Q.In the Young's double slit experiment, when a monochromatic light is used, fringe width obtained is 1 mm . If the wave length is halved and the slit width is doubled, what will be the new fringe width? (A) 1 mm (B) 0.25 mm (C) 1.25 mm (D) 0.5 mm

   ›Reveal solutionSolution

   The fringe width in Young’s double-slit experiment is given by $\beta = \frac{\lambda D}{d}$. Halving the wavelength halves $\beta$; doubling the slit width halves $\beta$ again, so the new fringe width is $\frac{1}{4}$ of the original, i.e., $0.25\ \text{mm}$. The correct option is (B).

   The key concept here is the fringe width formula in Young’s double-slit experiment. Fringe width $\beta$ is the distance between consecutive bright (or dark) fringes on the screen. It depends directly on the wavelength $\lambda$ and the distance to the screen $D$, and inversely on the slit separation $d$. The problem changes $\lambda$ and $d$ independently, so we can find the new $\beta$ by scaling the original value.

   Let’s work through it step by step:

   1. Recall the formula For a double-slit setup with slit separation $d$, screen distance $D$, and wavelength $\lambda$, the fringe width is

   $\beta = \frac{\lambda D}{d}.$

   Here, $D$ is not mentioned, so we assume it stays constant.

   2. Original conditions Given: $\beta_1 = 1\ \text{mm}$, with wavelength $\lambda_1$ and slit separation $d_1$. So

   $\beta_1 = \frac{\lambda_1 D}{d_1} = 1\ \text{mm}.$

   3. New conditions Wavelength is halved: $\lambda_2 = \frac{\lambda_1}{2}$. Slit width is doubled — but careful: “slit width” in Young’s experiment usually means the width of each slit, not the separation between them. However, the fringe width formula uses the slit separation $d$ (distance between centers of the two slits). The problem likely means the slit separation is doubled (since changing individual slit width affects intensity, not fringe spacing). This is a classic trick: many students confuse “slit width” with “slit separation.” In standard YDSE problems, “slit width” often refers to the separation $d$. We’ll interpret it as $d_2 = 2d_1$. …

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9. COMEDK 2025Set 2025-E1 markMCQ
   Q.The interference pattern is obtained with two coherent light sources of intensity ratio $9: 1$. The ratio of $\frac{I_{\operatorname{MAX}}+I_{\operatorname{MIN}}}{I_{\operatorname{MAX}}-I_{\operatorname{MIN}}}$ is $\frac{\alpha}{\beta}$. The values of $\alpha$ and $\beta$ are: (A) 5 and 3 (B) 3 and 1 (C) 1 and 9 (D) 9 and 1

   ›Reveal solutionSolution

   The key idea is to express the maximum and minimum intensities in terms of the individual intensities, then simplify the given ratio. The result is $\frac{5}{3}$, so $\alpha = 5$ and $\beta = 3$, corresponding to option (A).

   Concept and Intuition

   When two coherent light sources interfere, the resultant intensity depends on the phase difference. The maximum intensity occurs when the waves are in phase (constructive interference), and the minimum when they are exactly out of phase (destructive interference). The given ratio of intensities of the two sources is $9:1$, meaning one is nine times stronger than the other. The expression $\frac{I_{\text{max}} + I_{\text{min}}}{I_{\text{max}} - I_{\text{min}}}$ simplifies to a neat form involving only the ratio of the amplitudes, making it independent of any scaling factor. This is a classic trick: the sum and difference of max and min intensities eliminate the constant background, leaving a simple ratio of amplitudes.

   Step-by-step solution

   1. Set up the intensities

      Let the intensities of the two sources be $I_1$ and $I_2$, with $I_1 : I_2 = 9 : 1$. So we can write $I_1 = 9I_0$ and $I_2 = I_0$ for some base intensity $I_0$.

      The amplitudes are proportional to the square roots of intensities: $A_1 = \sqrt{I_1} = 3\sqrt{I_0}$, $A_2 = \sqrt{I_2} = \sqrt{I_0}$.

   2. Maximum and minimum intensities

      For two coherent sources, the resultant intensity is

   $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \delta$

   where $\delta$ is the phase difference.

   • Maximum ($\cos \delta = 1$):

   $I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} = 9I_0 + I_0 + 2\sqrt{9I_0 \cdot I_0} = 10I_0 + 2 \cdot 3I_0 = 16I_0$

   • Minimum ($\cos \delta = -1$):

   $I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2} = 10I_0 - 6I_0 = 4I_0$

   3. Compute the required ratio

   $\frac{I_{\text{max}} + I_{\text{min}}}{I_{\text{max}} - I_{\text{min}}} = \frac{16I_0 + 4I_0}{16I_0 - 4I_0} = \frac{20I_0}{12I_0} = \frac{20}{12} = \frac{5}{3}$

   So $\alpha = 5$ and $\beta = 3$. …

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10. COMEDK 2025Set 2025-E1 markMCQ
    Q.The wavelength of a monochromatic light which is used in single slit diffraction is 800 nm . The width of the single slit for which the first minimum appears at $\theta=45^{\circ}$ on the screen will be: (A) $1.13 \mu \mathrm{~m}$ (B) $1.23 \mu \mathrm{~m}$ (C) $2.13 \mu \mathrm{~m}$ (D) $1.3 \mu \mathrm{~m}$

    ›Reveal solutionSolution

    The first minimum in single‑slit diffraction occurs when the path difference across the slit equals one wavelength. Using the condition $a \sin\theta = \lambda$ with $\lambda = 800\ \text{nm}$ and $\theta = 45^\circ$, the slit width is found to be about $1.13\ \mu\text{m}$, which corresponds to option (A).

    The key idea is that in single‑slit diffraction, the first minimum is caused by destructive interference between rays from the top and bottom of the slit. The condition is $a \sin\theta = m\lambda$ with $m = \pm 1$ for the first minimum. Here we simply solve for $a$.

    1. Recall the condition for minima in single‑slit diffraction For a slit of width $a$, the first minimum occurs when the path difference between a ray from the top edge and a ray from the bottom edge equals exactly one wavelength. This gives

    $a \sin\theta = \lambda$

    where $\theta$ is the angle from the central axis to the first minimum.

    2. Insert the given values We have $\lambda = 800\ \text{nm} = 800 \times 10^{-9}\ \text{m}$ and $\theta = 45^\circ$.

    $a \sin 45^\circ = 800 \times 10^{-9}$

    Since $\sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071$, we get

    $a \times 0.7071 = 800 \times 10^{-9}$

    3. Solve for $a$ $a = \frac{800 \times 10^{-9}}{0.7071} \approx 1.131 \times 10^{-6}\ \text{m}$ …

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11. COMEDK 2025Set 2025-E1 markMCQ
    Q.Young's double slit experiment is first done in air and then in a medium of refractive index $\mu$. If the 7 th dark fringe in the medium lies where the 4 th bright fringe is in air, then the value of $\mu$ is : (A) 1.654 (B) 1.389 (C) 1.875 (D) 1.768

    ›Reveal solutionSolution

    Immersing the whole set-up in the medium shrinks the fringe pattern by the factor $\mu$; equating the 7th dark fringe of the medium pattern to the 4th bright fringe of the air pattern gives $\mu = \dfrac{15}{8} = 1.875$.

    Setup

    In Young's double slit experiment the fringe spacing is $\beta = \dfrac{\lambda D}{d}$. When the apparatus is placed in a medium of refractive index $\mu$ the wavelength shrinks to $\lambda/\mu$, so the fringe spacing becomes

    $\beta' = \frac{\lambda D}{\mu\, d}.$

    Locate the two fringes

    • 4th bright fringe in air: $\;y_B = 4\beta = \dfrac{4\lambda D}{d}$.
    • 7th dark fringe in the medium: the destructive path difference for this minimum is $\tfrac{15}{2}\lambda$, so $\;y_D = \dfrac{15}{2}\,\beta' = \dfrac{15\,\lambda D}{2\mu\, d}.$

    Coincidence condition $\;y_D = y_B$: …

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12. COMEDK 2025Set 2025-M1 markMCQ
    Q.If E is the amplitude of the electric field of the waves starting from the slits in a double slit experiment and $\theta$ is the phase difference between the waves reaching a point on the screen, the ratio of the amplitude of the resultant electric field at that point on the screen to the amplitude at one of the slits is (A) $\cos (\theta)$ (B) $2 \cos (\theta)$ (C) $\cos \left(\frac{\theta}{2}\right)$ (D) $2 \cos \left(\frac{\theta}{2}\right)$

    ›Reveal solutionSolution

    The resultant amplitude from two coherent sources with equal amplitude $E$ and phase difference $\theta$ is $2E\cos(\theta/2)$, so the ratio to the amplitude at one slit ($E$) is $2\cos(\theta/2)$. The correct option is (D).

    Concept and intuition

    In a double-slit experiment, each slit acts as a source of coherent waves (same frequency, constant phase difference). When two waves of equal amplitude $E$ meet at a point on the screen, they superpose. The resultant amplitude depends on the phase difference $\theta$ between them. The key is to use the principle of superposition and the geometry of phasor addition: two vectors of equal length $E$ with an angle $\theta$ between them add to a resultant whose magnitude is $2E\cos(\theta/2)$. This is a standard result from vector addition or from the trigonometric identity for the sum of two sine waves.

    Step-by-step reasoning

    1. Set up the superposition Let the electric fields from the two slits at the point on the screen be

    $E_1 = E \sin(\omega t), \quad E_2 = E \sin(\omega t + \theta)$

    where $E$ is the amplitude from one slit, $\omega$ is the angular frequency, and $\theta$ is the phase difference due to path difference.

    2. Add the waves The resultant field is

    $E_{\text{res}} = E \sin(\omega t) + E \sin(\omega t + \theta)$

    Use the sum-to-product identity:

    $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

    Here $A = \omega t$, $B = \omega t + \theta$, so

    $E_{\text{res}} = 2E \sin\left(\omega t + \frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)$

    3. Identify the resultant amplitude The amplitude is the coefficient of the sine term (the maximum value of the oscillation): $\text{Resultant amplitude} = 2E \left|\cos\left(\frac{\theta}{2}\right)\right|$ …

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