Question 1

A multiple slit diffraction grating has a slit separation of 2.00 * 10 ^- ⁶ m. Find the wavelength of the monochromatic light that will have its second order (m = 2) bright fringe diffracted through an angle of 43.4°. (1 nm = 10 ^- ⁹ m)

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Question 2

An individual's eye pupil changes from a diameter of 1.5 mm to 3.5 mm as the illumination is increased. By what factor does the minimum angle of resolution change?

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Question 3

Diffraction grating #1 has twice the lines/cm that diffraction grating #2 does. When used with a certain wavelength of light, both gratings give first order (m = 1) maxima. Which of the following statements is true for the same wavelength of light.

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Question 4

A wavelength of 573 nm yields a first order (m = 1) maximum at 40° with a grating. At what angle will the second order (m = 2) maximum appear for this wavelength?

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Question 5

A microscope has an objective lens with an aperture diameter 0.60 cm. A monochromatic light source of wavelength 580 nm is used to illuminate the object. It is determined that the minimum angle of resolution is 1.18 *10^-⁴rad. If the present lens were replaced by one with an aperture of diameter 1.0 cm, what would the minimum angle of resolution now become? (1 nm = 10^-⁹ m)

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Question 6

If different filters are used with an astronomical telescope, which of the following would give the best resolution?

Accepted Answer

Shorter wavelengths, like blue light, provide better resolution due to less diffraction compared to longer wavelengths like red.

Question 7

At what angle will the second order (m = 2) maximum occur for a wavelength of 450 nm using a diffraction grating with 10,000 lines per cm?

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Question 8

Find the radius of a star image formed on the retina of the eye if the aperture diameter (the pupil) at night is 0.70 cm and the length of the eye is 3.0 cm. Assume the wavelength of starlight in the eye is 600 nm. (1 nm = 10 ^- ⁹m)

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The answer of Find the radius of a star image...

Question 9

In a single-slit diffraction experiment, the angular width of the central maximum is 0.16°. The width of the slit is then doubled as is the distance to the screen. What is the resulting angular width of the central maximum?

Accepted Answer

The angular width of the central maximum is given by:

θ = λ / a

where λ is the wavelength of light and a is the width of the slit.

Since the angular width of the central maximum is 0.16°, we can use this equation to find the value of λ/a.

0.16° = λ / a

For simplicity, we can assume that the wavelength of light is constant throughout the experiment. Therefore, λ/a is also constant.

When the width of the slit is doubled, the new value of a becomes 2a. Similarly, the new distance to the screen becomes 2d.

The new angular width of the central maximum is given by:

θ' = λ / 2a

Substituting 2a for a in the above equation, we get:

θ' = λ / (2a) = (1/2) (λ / a) = (1/2) θ

Therefore, the new angular width of the central maximum is half the original value, which is:

θ' = (1/2) (0.16°) = 0.080°

Hence, the answer is B.

Question 10

Light with a wavelength of 450 nm shines through a lens with an aperture diameter of 0.85 cm. Use Rayleigh's criterion to determine the limiting angle of resolution. (1 nm = 10^-⁹m)

Accepted Answer

Rayleigh's criterion states that the limiting angle of resolution is given by:$$	heta = 1.22 \frac{\lambda}{D}$$where:$$	heta$$ is the limiting angle of resolution$$\lambda$$ is the wavelength of light$$D$$ is the diameter of the aperturePlugging in the given values, we get:$$	heta = 1.22 \frac{450 	imes 10^{-9} 	ext{ m}}{0.85 	imes 10^{-2} 	ext{ m}} = 6.5 	imes 10^{-5} 	ext{ rad}$$Therefore, the correct choice is A.

Question 11

A diffraction grating has 2500 lines/cm. What is the slit separation?

Accepted Answer

The slit separation is the reciprocal of the number of lines per centimeter. So, the slit separation is 1/2500 cm = 4.0 x 10^-4 cm = 400 nm.

Question 12

What is the highest order (m value) maximum for wavelength 450 nm that can be obtained with a grating with 300 lines per mm?

Accepted Answer

The highest-order maximum (m value) can be calculated using the grating equation $d\sin(\theta) = m\lambda$, where $d$ is the grating spacing, $\lambda$ is the wavelength, and $m$ is the order of the maximum. The grating spacing $d$ is the inverse of the number of lines per mm, so for 300 lines per mm, $d = \frac{1}{300 \text{ mm}^{-1}} = \frac{1}{300000 \text{ m}^{-1}} = 3.33 \times 10^{-6} \text{ m}$. To find the highest possible order, we need the maximum value of $\sin(\theta)$, which is 1 (when $\theta = 90^\circ$). Thus, $m = \frac{d}{\lambda}$. Substituting the given values, $m = \frac{3.33 \times 10^{-6} \text{ m}}{450 \times 10^{-9} \text{ m}} = 7.4$. Since $m$ must be an integer, the highest possible order is 7.

Question 13

A laser shines 630-nm light through a grating producing a first order (m = 1) maximum at an angle of 30°. A second laser produces a first order maximum with the same grating at 20°. What is the wavelength emitted by the second laser?

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Question 14

What is the highest order (the m value) of the complete visible spectrum (390 nm to 710 nm) that can be obtained with a grating having 300 lines/mm?

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Question 15

A microscope has an objective lens with an aperture of diameter 0.60 cm where a monochromatic light source of wavelength 580 nm is used to illuminate the object. It is determined that the minimum angle of resolution is 1.18 *10 ^- ⁴ rad. If the illuminating source were replaced by a source of wavelength 700 nm, what would the minimum angle of resolution now become? (1 nm = 10^-⁹m)

Accepted Answer

The minimum angle of resolution is given by the formula θ = 1.22 * (λ/D), where λ is the wavelength and D is the diameter of the aperture. When the wavelength changes from 580 nm to 700 nm, the angle of resolution increases proportionally. Therefore, the new angle is (700/580) * 1.18 x 10^-4 rad, which is approximately 1.4 x 10^-4 rad.

Question 16

In a single-slit diffraction experiment, the width of the central maximum is 6.0 mm. The width of the slit is then doubled as is the distance to the screen. What is the resulting width of the central maximum?

Accepted Answer

The width of the central maximum is directly proportional to the width of the slit and inversely proportional to the distance to the screen. Doubling the width of the slit will double the width of the central maximum, but also doubling the distance to the screen will reduce the width by a factor of 2. Therefore, the resulting width of the central maximum will be the same as before, which is 6.0 mm.

Question 17

At what angle will the highest order (m value) maximum appear for a wavelength 400 nm using a grating with 600 lines per mm?

Accepted Answer

The grating equation is given by mλ = d(sinθ + sinφ), where m is the order of diffraction, λ is the wavelength of light, d is the spacing between the lines of the grating, θ is the angle of incidence, and φ is the angle of diffraction. Solving for φ and plugging in the given values, we get sinφ = (mλ/d) - sinθ. For the highest order maximum, we take m = N (where N is the total number of lines illuminated by the incident light) = (400 nm * 600 lines per mm) = 240. Plugging this value into the equation and solving for θ, we get θ = 73.99°, which is closest to choice C (74°).

Question 18

A diffraction grating with 10,000 lines/cm will exhibit the first order (m = 1) maximum for light of wavelength 264 nm at what angle? (1 nm = 10 ^- ⁹ m)

Accepted Answer

The correct answer is A because the first order (m = 1) maximum for light of wavelength 264 nm will exhibit at an angle of 15.3°.

Question 19

The 2.4-m diameter Hubble space telescope has been placed into Earth orbit by the space shuttle. Assuming a wavelength of 710 nm, what angular resolution could this telescope achieve by Rayleigh's criterion? (1 nm = 10^-⁹m)

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Question 20

A binary star system in the constellation Orion has an angular separation between the stars of 10^-⁵radians. Assuming a wavelength of 700 nm, what is the smallest aperture (diameter) telescope that will just resolve the two stars? (1 nm = 10^-⁹m)

Accepted Answer

The Rayleigh criterion for resolving two point sources is given by θ = 1.22 * (λ/D), where θ is the angular separation, λ is the wavelength, and D is the diameter of the aperture. Solving for D gives D = 1.22 * (λ/θ). Substituting λ = 700 nm = 700 x 10^-9 m and θ = 10^-5 radians, we find D ≈ 0.085 m or 8.5 cm.

Quiz 25: Wave Optics

A multiple slit diffraction grating has a slit separation of 2.00 * 10 - 6 m. Find the wavelength of the monochromatic light that will have its second order (m = 2) bright fringe diffracted through an angle of 43.4°. (1 nm = 10 - 9 m)