Physics NCERT Exemplar Solutions Class 12th Chapter Ten Wave Optics Class 12th Physics

Orange light of wavelength $6000 * 10^{- 10} m$ illuminates a single slit of width $0.6 * 10^{- 4} m$ . the maximum possible number of diffraction minima produced on both sides of the central maximum is

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Alok Kumar Singh | Contributor-Level 10

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Sol. Angular momentum conservation:

$\begin{array}{r} \Rightarrow I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{f} \\ \Rightarrow \frac{{M R}^{2}}{2} ω_{o} = (\frac{{M R}^{2}}{2} + \frac{{M R}^{2}}{8}) ω_{f} \\ \Rightarrow ω_{f} = \frac{4}{5} ω_{o} \\ \Rightarrow {K E}_{final} = \frac{1}{2} (I_{1} + I_{2}) ω_{f}^{2} = \frac{{M R}^{2} ω^{2}}{5} \\ \Rightarrow K E_{initial} = \frac{1}{2} I, ω_{0}^{2} = \frac{{M R}^{2} ω^{2}}{4} \\ \Rightarrow % loss \Rightarrow 20 % . \end{array}$

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The angle between the plane of vibration and plane of polarization is

Alok Kumar Singh

The angle between the plane of vibration and plane of polarization is 90°.

A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wave-train of wavelength 6 cm is produced at the lower end of the rope. What is the wavelength of the wave train (in cm) when it reaches the top of the rope?


Alok Kumar Singh

At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm

At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?

Since frequency (f) remains the same:
f = v?

In the Young's double slit experiment, the distance between the slits varies in time as d(t) = d? + a? sin(ωt); where d?, ω and a? are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
= λD [ (d? + a? ) - (d? - a? ) / (d? ² - a? ²) ]
= 2λDa? / (d? ² - a? ²)

An unpolarised light beam of intensity 2l₀ is passed through a polaroid P and then through another polaroid Q which is oriented is such a way that its passing axis makes an angle of 30° relative to that of P. The intensity of the emergent light is

$I = I_{0} c o s^{2} 30 °$

$= I_{0} {(\frac{\sqrt[]{3}}{2})}^{2} = \frac{3}{4} I_{0}$

In Young’s double slit experiment the two slits are 0.6mm distance apart. Interference pattern is observed on a screen at a distance 80cm from the slits. The first dark fringe is observed on the screen directly opposite to one of the slits. The wavelength of light will be__________nm.

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\Rightarrow \frac{d y}{D} = (2 n + 1) \frac{λ}{2}$ [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{d y}{D} = \frac{λ}{2}$

$λ = \frac{2 d}{D} y$

$= \frac{2 d}{D} \times \frac{d}{2} [y = \frac{d}{2}, G i v e n$ first dark fringe is observed on the screen directly opposite to one of the slits]

$λ = \frac{2 \times 0.6 \times 0.6}{800 \times 2}$

$λ = 450 m m$

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Physics NCERT Exemplar Solutions Class 12th Chapter Ten 2025

Physics NCERT Exemplar Solutions Class 12th Chapter Ten 2025

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