When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :
- \(60^\circ\)
- \(30^\circ\)
- \(90^\circ\)
- \(45^\circ\)
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to understand the concept of polaroid sheets and how light intensity varies when a polaroid sheet is rotated between two crossed polaroids.
When two Polaroids are crossed (90 degrees apart), ideally, no light should pass through. This is because the plane of polarization is perpendicular to each other, blocking any light.
Inserting a third polaroid sheet at an angle between the two crossed polaroids will allow some light to pass through due to the mechanism of polarization, which follows Malus's Law.
Malus's Law: The intensity \(I\) of light transmitted through a Polaroid is given by:
\(I = I_0 \cos^2\theta\)
where \(I_0\) is the initial intensity of the light, and \(\theta\) is the angle between the light's initial polarization direction and the axis of the Polaroid.
The problem asks for the angle at which the transmitted light intensity is maximum when a Polaroid sheet is rotated between two crossed Polaroids. The maximum transmitted intensity occurs when the intervening Polaroid is at an angle of \(45^\circ\) to each of the crossed polaroids.
Let's apply this concept:
- \(\theta = 45^\circ\) between the first Polaroid and the middle sheet.
- The remaining rotation, \(90^\circ - 45^\circ = 45^\circ\), is between the middle sheet and the last Polaroid.
Therefore, for maximum transmitted light intensity, the middle Polaroid should be rotated by \(45^\circ\).
Hence, the correct answer is \(45^\circ\).
Approach Solution -2
Let \( I_0 \) be the intensity of unpolarized light incident on the first polaroid.
- The transmitted light intensity after passing through the first polaroid is:
\(I_1 = \frac{I_0}{2}.\)
- The intensity after passing through the second polaroid at an angle \( \theta \) is:
\(I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta.\)
- Similarly, for the third polaroid rotated by \( 90^\circ - \theta \), the transmitted intensity is:
\(I_3 = I_2 \cos^2 (90^\circ - \theta) = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta.\)
- The intensity will be maximum when \( \sin 2\theta = 1 \), which occurs at \( \theta = 45^\circ \).
The Correct answer is: \(45^\circ\)
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