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In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0.2 mm, the angular width of central maximum is $\alpha \times 10^{-2}$ degrees. The value of $\alpha$ is _________.
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Answer: 36
Key idea: Single Slit Diffraction - Angular Width of Central Maximum
For single slit diffraction, first minimum occurs at $a \sin \theta = \lambda$, where $a$ is slit width
For small angles, $\theta \approx \lambda/a = \frac{628 \times 10^{-9}}{2 \times 10^{-4}} = 3.14 \times 10^{-3}$ radians
Angular width of central maximum = $2\theta = 2 \times 3.14 \times 10^{-3} = 6.28 \times 10^{-3}$ radians
In a Young's double slit experiment, the intensity at some point on the screen is found to be $\frac{3}{4}$ times of the maximum of the interference pattern. The path difference between the interfering waves at this point is $\frac{\lambda}{x}$ where $\lambda$ is wavelength of the incident light. The value of $x$ is ________.
In YDSE with equal intensity sources, the intensity formula is $I = 4I_0\cos^2(\phi/2)$ where $I_0$ is individual source intensity and $\phi$ is phase difference
Maximum intensity occurs at $\phi = 0$, giving $I_{\text{max}} = 4I_0$
Given $I = \frac{3}{4}I_{\text{max}}$, substitute: $4I_0\cos^2(\phi/2) = \frac{3}{4} \times 4I_0$
Solving: $\cos^2(\phi/2) = \frac{3}{4}$, so $\cos(\phi/2) = \frac{\sqrt{3}}{2}$, giving $\phi/2 = \frac{\pi}{6}$ or $\phi = \frac{\pi}{3}$
Use relation between phase and path difference: $\phi = \frac{2\pi}{\lambda}\Delta$
A slit of width $a$ is illuminated by light of wavelength $\lambda$. The linear separation between $1^{st}$ and $3^{rd}$ minima in the diffraction pattern produced on a screen placed at a distance $D$ from the slit system is ________.
(1) $\frac{D\lambda}{a}$
(2) $1.5\frac{D\lambda}{a}$
(3)
(4)
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Answer: Option (2Dλ/a (likely option 3 or 4, as options 1 and 2 don't match))
Key idea: Single Slit Diffraction - Position of Minima
For single slit diffraction, minima occur at $a \sin\theta = n\lambda$ where $n = 1, 2, 3, ...$
For small angles, $\sin\theta \approx \frac{y}{D}$, so position of $n^{th}$ minimum is $y_n = \frac{n\lambda D}{a}$
Position of 1st minimum: $y_1 = \frac{\lambda D}{a}$
Position of 3rd minimum: $y_3 = \frac{3\lambda D}{a}$
Linear separation: $\Delta y = y_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2D\lambda}{a}$
An unpolarized light is incident on the plane interface of air-dielectric medium shown in figure. If the incident angle is equal to Brewster angle, identify the expression representing reflected wave. [Figure: Unpolarized incident light from air hitting air-medium interface at an angle, with coordinate system showing x, y, z axes]
An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at $30°$ and $90°$, respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at $60°$ with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is ______.
(1) $3/4$
(2) $4/3$
(3) $9/4$
(4) $4/9$
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Answer: Option (3)
Key idea: Malus's Law and Polarization of Light
**Without third polarizer:** Unpolarized light (intensity $I_0$) passes through first polarizer at $30°$: $I_1 = I_0/2$
Light at $30°$ passes through second polarizer at $90°$. Angle difference = $60°$. Using Malus's Law: $I_{out} = \frac{I_0}{2} \cos^2(60°) = \frac{I_0}{8}$
**With third polarizer:** After first polarizer: $I_1 = I_0/2$ (polarized at $30°$)
Through third polarizer at $60°$. Angle difference = $30°$: $I_2 = \frac{I_0}{2} \cos^2(30°) = \frac{3I_0}{8}$
Through second polarizer at $90°$. Angle difference = $30°$: $I_{out} = \frac{3I_0}{8} \cos^2(30°) = \frac{9I_0}{32}$
Ratio = $\frac{9I_0/32}{I_0/8} = \frac{9I_0}{32} \times \frac{8}{I_0} = \frac{9}{4}$
In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of $7^{th}$ bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm, the thickness of mica sheet is $a \times 10^{-9}$ m. The value of $a$ is
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Answer: 5625
Key idea: Path difference introduced by transparent sheet in double slit experiment
When a mica sheet of thickness $t$ and refractive index $\mu$ is placed over one slit, additional path difference introduced is $(\mu - 1)t$
For central fringe to shift to $n^{th}$ bright fringe position: $(\mu - 1)t = n\lambda$
In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is $2.4$ μm. If the experiment is carried out in another medium having refractive index $1.2$, the fringe width will be ____ μm.
(1) $1.2$
(2) $2$
(3) $2.4$
(4) $2.88$
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Answer: Option (2)
Key idea: Effect of medium on fringe width in Young's double slit experiment
Recall that fringe width is given by $\beta = \frac{\lambda D}{d}$
When experiment is performed in a medium of refractive index $n$, wavelength changes: $\lambda_{\text{medium}} = \frac{\lambda_0}{n}$
The maximum intensity in a Young's double slit experiment is $I_o$. Distance between the slits ($d$) is 5$\lambda$, where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D=10d$ is ________.
(1) $\frac{I_0}{4}$
(2) $\frac{I_0}{2}$
(3) $I_0$
(4) $\frac{3I_0}{4}$
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Answer: Option (2)
Key idea: Young's Double Slit Experiment - Intensity at arbitrary point using path difference and phase difference
Given: $d = 5\lambda$, $D = 10d = 50\lambda$. Point P is directly opposite to slit S₁.
Calculate distances: $r_1 = D = 50\lambda$ and $r_2 = \sqrt{D^2 + d^2} = \sqrt{(50\lambda)^2 + (5\lambda)^2} = 5\lambda\sqrt{101}$
In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $\lambda/3$, where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda/6$. The ratio of intensities at points $A$ and $B$ is ______.
(1) 3
(2) 4
(3) 1/3
(4) 1/4
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Answer: Option (3)
Key idea: Interference intensity formula with path difference
For interference with equal intensity sources, intensity is given by $I = 4I_0\cos^2(\phi/2)$ where $\phi$ is the phase difference
Phase difference relates to path difference as $\phi = \frac{2\pi}{\lambda}\Delta$
At point A: $\phi_A = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$, so $I_A = 4I_0\cos^2(\frac{\pi}{3}) = 4I_0 \cdot \frac{1}{4} = I_0$
At point B: $\phi_B = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3}$, so $I_B = 4I_0\cos^2(\frac{\pi}{6}) = 4I_0 \cdot \frac{3}{4} = 3I_0$
An unpolarized light of intensity $I_o$ passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is $0°$ and intensity of light emerged from analyser is $\frac{3}{8}I_o$, the angle of rotation of the light by the solution with respect to analyser is ______ degrees.
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Answer: 30
Key idea: Malus's Law and Optical Activity
Unpolarized light of intensity $I_0$ becomes $\frac{I_0}{2}$ after passing through the polarizer
The optically active solution rotates the plane of polarization by angle $\theta$ without changing intensity
Since analyzer is at 0° (same as polarizer), the angle between rotated polarization and analyzer is $\theta$
In a Young double slit experiment, the wavelength of incident light is $6000$ Å, the separation between slits $S_1$ and $S_2$ is $5$ cm and the distance between slits plane and screen is $50$ cm, as shown in the figure below. If the resultant intensity at $P$ is equal to the intensity due to individual slits, the path difference between interfering waves is _______ Å. [Figure: Diagram showing Young's double slit setup with source S emitting light of wavelength λ = 6000Å, two slits S₁ and S₂ separated by 5 cm, a screen at distance 50 cm from the slits, and point P on the screen with point O' on the perpendicular from the slits to the screen]
(1) 4000
(2) 3000
(3) 2000
(4) 1000
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Interference of light waves and intensity distribution in Young's double slit experiment
FAQs
Frequently asked questions
How many questions come from Wave Optics in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 11 questions on it, about 1.2 per shift.
Where can I practise Wave Optics PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
