Consider sunlight incident on a slit of width 10 4 Å. The image seen through the slit shall (a) Be a fine sharp slit white in colour at the centre (b) A bright slit white at the centre diffusing to zero intensities at the edges (c) A bright slit white at the centre diffusing to regions of different colours (d) Only be diffused slit white in colour.

This is a Multiple Choice Questions as classified in NCERT ExemplarAnswer- ( a)Explanation- given width of slit is 10410-10m= 10-6Wavelength of sunlight varies from 4000A0 to 8000A0As the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear i.e., mixing of colours form white patch at the centre.

Ask & Answer Question

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

Consider sunlight incident on a slit of width 10 ⁴Å. The image seen through the slit shall
(a) Be a fine sharp slit white in colour at the centre
(b) A bright slit white at the centre diffusing to zero intensities at the edges
(c) A bright slit white at the centre diffusing to regions of different colours
(d) Only be diffused slit white in colour.

3 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- ( a)

Explanation- given width of slit is 10⁴10^-10m= 10^-6

Wavelength of sunlight varies from 4000A⁰ to 8000A⁰

As the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear i.e., mixing of colours form white patch at the centre.

0 Upvotes 0 Downvotes

Similar Questions for you

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? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
⇒ λ? = 2 * 6 cm = 12 cm

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 :

Raj Pandey

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

Payal Gupta

$I = I_{0} c o s^{2} 30 °$

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

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.

Vishal Baghel

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]

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

$λ = 450 m m$

In a Young's double slit experiment, the separation between the slits is $0.15 m m$ . In the experiment, a source of light of wavelength $589 n m$ is used and the interference pattern is observed on a screen kept $1.5 m$ away. The separation between the successive bright fringes on the screen is:

alok kumar singh

$\frac{Energy}{Volume} = \frac{{M L}^{2} T^{- 2}}{L^{3}} = ({M L}^{- 1} T^{- 2})$

The distance between two successive bright fringes is fringe width $(β)$ .

$β = \frac{λ D}{d} = \frac{589 \times 10^{- 9} \times 1.5}{0.15 \times 10^{- 3}} = 5.9 m m$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question