Question:
Interference pattern is observed at 'P' due to superposition of two rays coming out from a source 'S' as shown in the figure. The value of 'l' for which maxima is obtained at 'P' is: (R is perfect reflecting surface)
Interference pattern is observed at 'P' due to superposition of two rays coming out from a source 'S' as shown in the figure. The value of 'l' for which maxima is obtained at 'P' is: (R is perfect reflecting surface)
Show Hint
Find the reflected path length from the geometry, add the extra half-wavelength phase shift on reflection from the mirror, and set the effective path difference equal to an integral multiple of the wavelength.
Updated On: Nov 11, 2026
- \(l=\dfrac{2n\lambda}{\sqrt3-1}\)
- \(l=\dfrac{(2n-1)\lambda}{2(\sqrt3-1)}\)
- \(l=\dfrac{(2n-1)\lambda\sqrt3}{4(2-\sqrt3)}\)
- \(l=\dfrac{(2n-1)\lambda}{\sqrt3-1}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The direct ray travels SP = 2l. The reflected ray goes from S to the mirror at O and then to P. Taking the 30° angle as the angle made by the rays with the mirror (and with SP, which is parallel to the mirror), triangle SOP is isosceles with base angles 30° and half-base l.
SO = OP = l/cos 30° = 2l/\(\sqrt3\).
Reflected path = 4l/\(\sqrt3\).
Geometrical path difference = \(\frac{4l}{\sqrt3}-2l=\frac{2l(2-\sqrt3)}{\sqrt3}\).
Reflection from the perfectly reflecting surface adds a phase change of \(\pi\), i.e. an extra \(\lambda/2\).
For maxima: \(\frac{2l(2-\sqrt3)}{\sqrt3}+\frac{\lambda}{2}=n\lambda\), so \(\frac{2l(2-\sqrt3)}{\sqrt3}=\frac{(2n-1)\lambda}{2}\).
Hence \(l=\frac{(2n-1)\lambda\sqrt3}{4(2-\sqrt3)}\).
Answer: (3) \(l=\dfrac{(2n-1)\lambda\sqrt3}{4(2-\sqrt3)}\)
SO = OP = l/cos 30° = 2l/\(\sqrt3\).
Reflected path = 4l/\(\sqrt3\).
Geometrical path difference = \(\frac{4l}{\sqrt3}-2l=\frac{2l(2-\sqrt3)}{\sqrt3}\).
Reflection from the perfectly reflecting surface adds a phase change of \(\pi\), i.e. an extra \(\lambda/2\).
For maxima: \(\frac{2l(2-\sqrt3)}{\sqrt3}+\frac{\lambda}{2}=n\lambda\), so \(\frac{2l(2-\sqrt3)}{\sqrt3}=\frac{(2n-1)\lambda}{2}\).
Hence \(l=\frac{(2n-1)\lambda\sqrt3}{4(2-\sqrt3)}\).
Answer: (3) \(l=\dfrac{(2n-1)\lambda\sqrt3}{4(2-\sqrt3)}\)
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
Top JEE Main Wave optics Questions
- When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :
- In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m. the minimum distance between the slits S1 and S2 is ______ mm.
- A monochromatic light of wavelength 6000Å is incident on the single slit of width 0.01 mm. If the diffraction pattern is formed at the focus of the convex lens of focal length 20 cm, the linear width of the central maximum is :
- Monochromatic light of wavelength \( 500 \, \text{nm} \) is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index \( = 1.5 \)), the central maximum is shifted to a position previously occupied by the 4\textsuperscript{th} bright fringe. The thickness of the glass plate is ___________ \( \mu \text{m} \).
- Two wavelengths \( \lambda_1 \) and \( \lambda_2 \) are used in Young's double slit experiment. \( \lambda_1 = 450 \, \text{nm} \) and \( \lambda_2 = 650 \, \text{nm} \). The minimum order of fringe produced by \( \lambda_2 \), which overlaps with the fringe produced by \( \lambda_1 \), is \( n \). The value of \( n \) is _____.
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions