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34. Wave Optics

Single Slit Diffraction

Physics

34. Wave Optics

Single Slit Diffraction

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4:19 minutes

Problem 6

Textbook Question

Monochromatic light of wavelength 633 nm falls on a slit. If the angle between the first bright fringes on either side of the central maximum is 32°, estimate the slit width.

Verified step by step guidance

Identify the relevant formula for single-slit diffraction. The angular position of the minima in a single-slit diffraction pattern is given by:

a \sin(\theta) = m \lambda

, where

a

is the slit width,

\theta

is the angle to the m-th minimum,

m

is the order of the minimum, and

\lambda

is the wavelength of the light.

Recognize that the problem provides the total angle between the first bright fringes on either side of the central maximum as 32°. This means the angle to the first minimum on one side of the central maximum is half of this value:

\theta = \frac{32^\circ}{2} = 16^\circ

Substitute the known values into the formula. For the first minimum,

m = 1

\lambda = 633 \ \text{nm} = 633 \times 10^{-9} \ \text{m}

, and

\theta = 16^\circ

. The equation becomes:

a \sin(16^\circ) = 1 \cdot 633 \times 10^{-9}

Rearrange the equation to solve for the slit width

a

a = \frac{633 \times 10^{-9}}{\sin(16^\circ)}

Use a calculator to compute

\sin(16^\circ)

and substitute it into the equation to find the value of

a

. Ensure the angle is in degrees when calculating the sine function.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Diffraction

Diffraction is the bending of waves around obstacles and the spreading of waves when they pass through narrow openings. In the context of light, diffraction patterns are created when light waves encounter a slit, leading to the formation of bright and dark fringes on a screen. The extent of diffraction depends on the wavelength of the light and the size of the slit.

Young's Double Slit Experiment

Young's Double Slit Experiment demonstrates the wave nature of light through the interference pattern created by two closely spaced slits. The positions of the bright and dark fringes are determined by the path difference between light waves emanating from the slits. This experiment is foundational in understanding how light behaves as a wave, particularly in terms of constructive and destructive interference.

Slit Width and Wavelength Relationship

The relationship between slit width, wavelength, and the angle of diffraction is described by the formula for single-slit diffraction. The width of the slit affects the angle at which the first bright fringe appears; narrower slits produce wider diffraction patterns. This relationship is crucial for calculating the slit width when given the wavelength of light and the angular separation of the fringes.

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Related Practice

Textbook Question

Light from a helium-neon laser (λ = 633 nm) passes through a circular aperture and is observed on a screen 4.0 m behind the aperture. The width of the central maximum is 2.5 cm. What is the diameter (in mm) of the hole?

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Textbook Question

The electrons in a cathode-ray tube are accelerated through a 250 V potential difference and then shot through a 33-nm-diameter circular aperture. What is the diameter of the bright spot on an electron detector 1.5 m behind the aperture?

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Textbook Question

Monochromatic light falls on a slit that is 2.60 x 10⁻³ mm wide. If the angle between the first dark fringes on either side of the central maximum is 29.0° (dark fringe to dark fringe), what is the wavelength of the light used?

391

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Textbook Question

Light of wavelength 580 nm falls on a slit that is 3.50 x 10⁻³ mm wide. Estimate how far the first brightest diffraction fringe is from the strong central maximum if the screen is 10.0 m away.

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Textbook Question

(a) For a given wavelength λ, what is the minimum slit width for which there will be no diffraction minimum? (b) What is the minimum slit width so that no visible light exhibits a diffraction minimum?

630

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Textbook Question

A single slit 0.10 mm wide is illuminated by 450-nm light. What is the width of the central maximum in the diffraction pattern on a screen 5.0 m away?

355

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Textbook Question

If you double the width of a single slit, the intensity of the light passing through the slit is doubled. (a) Show, however, that the intensity at the center of the screen increases by a factor of 4. (b) Explain why this does not violate conservation of energy.

333

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Textbook Question

(a) Explain why the secondary maxima in the single-slit diffraction pattern do not occur precisely at β/2 = (m + 1/2)π where m = 1, 2, 3, ... .

(b) By differentiating Eq. 35–7 with respect to β show that the secondary maxima occur when β/2 satisfies the relation tan(β/2) = β/2.

(c) Carefully and precisely plot the curves y = β/2 and y = tan β/2. From their intersections, determine the values of β for the first and second secondary maxima. What is the percent difference from β/2 = (m + 1/2)π?

313

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