Question:
An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at \(30^\circ\) and \(90^\circ\), respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at \(60^\circ\) with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is:
An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at \(30^\circ\) and \(90^\circ\), respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at \(60^\circ\) with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is:
Updated On: Apr 10, 2026
- 3/4
- 4/3
- 9/4
- 4/9
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
When unpolarized light passes through a polarizer, its intensity is reduced to half. When polarized light passes through subsequent polarizers, the intensity is governed by Malus' Law, which states that the transmitted intensity is proportional to the square of the cosine of the angle between the transmission axes.
Step 2: Key Formula or Approach:
1. Intensity after first polarizer: \( I_1 = \frac{I_0}{2} \).
2. Malus' Law: \( I = I_{\text{incident}} \cos^2 \theta \).
Step 3: Detailed Explanation:
Let the initial intensity of unpolarized light be \( I_0 \). After the first polarizer at \( 30^\circ \), intensity is \( I_1 = \frac{I_0}{2} \).
Case 1: Without the third polarizer:
Angle between the axes of the two polarizers is \( \Delta \theta = 90^\circ - 30^\circ = 60^\circ \).
Output intensity \( I_{\text{without}} = I_1 \cos^2(60^\circ) = \frac{I_0}{2} \left( \frac{1}{2} \right)^2 = \frac{I_0}{8} \).
Case 2: With the third polarizer at \( 60^\circ \) in between:
Step i: Light passes from \( 30^\circ \) to \( 60^\circ \) polarizer. Angle diff = \( 30^\circ \).
\( I_2 = I_1 \cos^2(30^\circ) = \frac{I_0}{2} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3I_0}{8} \).
Step ii: Light passes from \( 60^\circ \) to \( 90^\circ \) polarizer. Angle diff = \( 30^\circ \).
\( I_{\text{with}} = I_2 \cos^2(30^\circ) = \frac{3I_0}{8} \times \frac{3}{4} = \frac{9I_0}{32} \).
Calculating the Ratio:
Ratio = \( \frac{I_{\text{with}}}{I_{\text{without}}} = \frac{9I_0/32}{I_0/8} = \frac{9}{32} \times \frac{8}{1} = \frac{9}{4} \).
Step 4: Final Answer:
The ratio of the output intensities with and without the third polarizer is 9/4.
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- 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:-
View More Questions
Top JEE Main Polarization Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- 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 ______.
View More Questions