Diffraction and the Wave Theory of Light MCQ Quiz in मल्याळम - Objective Question with Answer for Diffraction and the Wave Theory of Light - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

• Light is a form of energy that makes us the sensation of sight.
• Light is a form of Electromagnetic waves.
• The speed of light in the vacuum is 299,792,458 m/s and it is denoted with a small c.
• It does not need a medium to travel.
• Light travels in a straight line in a medium or vacuum (rectilinear propagation of light).

EXPLANATION:

• It can be reflected or absorbed wholly as in the case of opaque materials or partially as in the case of translucent materials and the amusing colour patterns are caused by the process of dispersion. Therefore option 1 is incorrect.
• Light travels in a straight line and bends when there is a change in the speed of light waves while travelling from one medium to another. Therefore option 2 is incorrect.
• The shadows are caused when the light source is blocked and eclipses are caused to the alignment in the position of Sun, Moon, and Earth.
• Light travels in a straight-line path.
  • When a small opaque object is kept in its path, the light tends to bend around the object and not walk in the same straight line. This is known as the diffraction of light. Therefore option 3 is incorrect.

Concept:

y = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{550 \times 10^{-9} \times 100 \times 10^{-2}}{0.2 \times 10^{-3}}\) = 275

∴ The distance of first order minima from the central maximum will be x × 10–5 m. The value of x is : 275

Concept:

Diffraction

• Diffraction is the phenomenon of light which describes the wave nature of light.
• Diffraction is the interference or bending of waves around the corners of an obstacle or through an aperture into the region of the geometrical shadow of the obstacle/aperture.
• In Young's double slit experiment, for maxima, path difference = nλ.

Angular spread

• Angular spread or Angular width is referred to as the length between two constructive dark or bright fringes.
• Angular spread is determined by a Sin θ = n λ,
• where a = Width of slit,
•  λ = Wavelength 

Calculation:

Here, a = 4 × 10², λ = 2 × 102

Since, a Sin θ = n λ 

⇒ Sin θ = nλ / a

⇒ Sin θ = (2 × 10²⁾ /( 4 × 10²) 

⇒ Sin θ = 1/2

⇒ Sin θ = 30°

The angular spread of the central maximum is: 2θ 

Thus the angular spread will be 60°

∴ The correct option is 3)

braggs law states

2d sinø = nλ

for first order n = 1

λ /d = 2 sin 32

for 2nd order n = 2

\(\sin {\theta _2} = \frac{{\lambda }}{d} = 2\sin {\theta _1} = 2\sin 32\)

2 sin 32 > 1 (which is never possible, hence no 2^nd order diffraction)

\( \dfrac{\sin \theta_{2}}{\sin \theta_{1}} = \dfrac{n_{2}}{n_{1}} \)

Hence, \( \sin \theta_{2} = 2 \sin 32 > 1 \)

Hence, a second order diffraction pattern doesn't exist in this case.

Calculation:

To determine the maximum number of intensity minima in the Fraunhofer diffraction pattern of a single slit, we use the condition for minima given by:

\(a \sin \theta = m \lambda\)

where;

a: is the slit width \(10 \, \mu \mathrm{m} = 10 \times 10^{-6} \, \mathrm{m}\)

\(\lambda\): is the wavelength of light \(0.630 \, \mu \mathrm{m} = 0.630 \times 10^{-6}m\).

m:  is the order of the minima \(m = \pm 1, \pm 2, \pm 3, \)

The maximum order m  can be found using:

\(m_{\text{max}} = \frac{a}{\lambda}.\)

Substituting the given values:

\(m_{\text{max}} = \frac{10 \times 10^{-6}}{0.630 \times 10^{-6}} \approx 15.87.\)

Since m must be an integer, the maximum order of minima is:

\(m = 15\)

Thus, option '4' is correct.

Explanation:

The angular width of the central maximum in the diffraction pattern of a single slit is determined by the formula:

\(\theta=2sin^{-1}(\frac{\lambda}{a})\)

where;

• a is the width of the slit.
• \(\lambda\) is the wavelength of light.

From the formula, the angular width depends on the wavelength of light (\(\lambda\)) and the slit width (a). However, it does not depend on the following:

• The frequency of light (because wavelength is related to frequency, but the dependence is implicit).
• The distance between the slit and the screen (because angular width is independent of this distance).

So the correct answer is: Distance between slit and source

Thus, option '3' is correct.

Concept:

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Explanation: 

The linear width of the central maximum on the screen is given by \( 2L\tan(\theta) \approx 2L\theta \) when θ" id="MathJax-Element-430-Frame" role="presentation" style="position: relative;" tabindex="0">θθ" id="MathJax-Element-139-Frame" role="presentation" style="position: relative;" tabindex="0">θθ θ $\theta$ is small, where L" id="MathJax-Element-431-Frame" role="presentation" style="position: relative;" tabindex="0">LL" id="MathJax-Element-140-Frame" role="presentation" style="position: relative;" tabindex="0">LL L $L$ is the distance from the slit to the screen.

Given:

• Slit width, a=0.1mm=0.1×10−3m" id="MathJax-Element-432-Frame" role="presentation" style="position: relative;" tabindex="0">a=0.1mm=0.1×10−3ma=0.1mm=0.1×10−3m" id="MathJax-Element-141-Frame" role="presentation" style="position: relative;" tabindex="0">a=0.1mm=0.1×10−3ma=0.1mm=0.1×10−3m a=0.1mm=0.1×10−3m $a = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m}$
• Linear width of central maxima, w=5mm=5×10−3m" id="MathJax-Element-433-Frame" role="presentation" style="position: relative;" tabindex="0">w=5mm=5×10−3mw=5mm=5×10−3m" id="MathJax-Element-142-Frame" role="presentation" style="position: relative;" tabindex="0">w=5mm=5×10−3mw=5mm=5×10−3m w=5mm=5×10−3m $w = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m}$
• Distance from the slit to the screen, L=50cm=0.5m" id="MathJax-Element-434-Frame" role="presentation" style="position: relative;" tabindex="0">L=50cm=0.5mL=50cm=0.5m" id="MathJax-Element-143-Frame" role="presentation" style="position: relative;" tabindex="0">L=50cm=0.5mL=50cm=0.5m L=50cm=0.5m $L = 50 \, \text{cm} = 0.5 \, \text{m}$

Using the approximation w≈2Lθ" id="MathJax-Element-435-Frame" role="presentation" style="position: relative;" tabindex="0">w≈2Lθw≈2Lθ" id="MathJax-Element-144-Frame" role="presentation" style="position: relative;" tabindex="0">w≈2Lθw≈2Lθ w≈2Lθ $w \approx 2L\theta$ :

\( \theta = \frac{w}{2L} = \frac{5 \times 10^{-3}}{2 \times 0.5} = 5 \times 10^{-3} \, \text{radians} \)

Now, using the formula θ=λa" id="MathJax-Element-436-Frame" role="presentation" style="position: relative;" tabindex="0">θ=λaθ=λa" id="MathJax-Element-145-Frame" role="presentation" style="position: relative;" tabindex="0">θ=λaθ=λa θ=λa $\theta = \frac{\lambda}{a}$ :

\( \lambda = \theta \times a = 5 \times 10^{-3} \times 0.1 \times 10^{-3} = 5 \times 10^{-7} \, \text{m} \)

The correct option is (3).

Concept:

• The wavelength of light is defined as “The distance between the two successive crests or troughs of the light wave”. It is denoted by the Greek letter lambda (λ)
• The visible spectrum is defined as the observable region of the electromagnetic wave which can be visible to the naked human eyes.
• In the electromagnetic spectrum, the visible spectrum ranges from the infrared region to the ultraviolet region.

Explanation:

• The wavelength of visible light ranges from 400 nm to 700 nm. 
• From option we can say that visible light ranges from 4 × 10^-7m - 8 × 10^-7m.
• This spectrum of visible light has numerous different colours and that too every colour with a different wavelength.
• The violet colour is said to have the shortest wavelength whereas the red colour is said to have the longest wavelength.

Additional Information

•  For the violet colour, the wavelength is between 380 to 450 nm and the frequency ranges from 668 to 789 THz.
• For the blue colour, the wavelength is between 450 to 495 nm and the frequency ranges from 606 to 668 THz.
• For the green colour, the wavelength is between 495 to 570 nm and the frequency ranges from 526 to 606 THz.
• For the yellow colour, the wavelength is between 570 to 590 nm and the frequency ranges from 508 to 526 THz.
• For the orange colour, the wavelength is between 590 to 620 nm and the frequency ranges from 484 to 508 THz.
• For the red colour, the wavelength is between 620 to 750 nm and the frequency ranges from 400 to 484 THz.

CONCEPT:

• When two light waves from different coherent sources meet together, then there is a distribution of energy in which one wave is disturbed by the other is called interference of light.

There are two conditions for the interference of light:

• Both the sources should be coherent (they must emit a light wave of constant phase difference.)
• The wave should be monochromatic (the light must be of same wavelength).

EXPLANATION:

• When white light fall on two different coloured filter- blue and red then it will emit two different light rays.
• One light ray will be red and another will be blue.
• Both the blue and red light have different wavelength, so their wavelengths are different and they are not monochromatic waves.
• Since for the interference to happen, there must be monochromatic waves.
• So there will not be any interference here in this case. Hence option 4 is correct.
• Since there is no interference, so no fringes will form.