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Butoxors [25]
3 years ago
10

Monochromatic light with a wavelength of 6.4 E -7 meter passes through two narrow slits, producing an interference pattern on a

screen 4.0 meters away. The first order bright band lines are 2.0 E -2 meters away from the central bright maxima. What is the distance between the slits?
1.3 E -4 m
8.0 E -6 m
3.2 E -9 m
7.8 E3 m
Physics
1 answer:
seropon [69]3 years ago
8 0

Answer:

The distance between the slits is given by  1.3 × 10^{-4} m

Given:

\lambda = 6.4 \times 10^{-7} m

D = 4 m

y = 2 \times 10^{-2} m

m = 1

To find:

distance between slits, d = ?

Formula used:

y = \frac{m \times \lambda \times D}{d}

y = distance of first bright band from central maxima

D = distance between screen and source

d = distance between slits

\lambda = wavelength

Solution:

distance of first bright band from central maxima is given by,

y = \frac{m \times \lambda \times D}{d}

y = distance of first bright band from central maxima

D = distance between screen and source

d = distance between slits

\lambda = wavelength

Thus,

d = \frac{m \times \lambda \times D}{y}

d = \frac{1 \times 6.4 \times 10^{-7} \times 4 }{2 \times 10^{-2} }

d = 1.28 × 10^{-4}

d = 1.3 × 10^{-4} m

The distance between the slits is given by  1.3 × 10^{-4} m

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Y_Kistochka [10]

Answer:

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5 0
3 years ago
What is the wavelength, in nm, of a photon with energy
lara31 [8.8K]

Answer:

(a)  λ = 4136 nm → infrared

(b) λ = 413.6 nm → visible light

(c) λ = 41.36 nm → ultraviolet

Explanation:

The wavelength of infrared is on the range of 700 nm to 1000000 nm

The wavelength of visible light is between 400 nm and 700 nm

The wavelength of ultraviolet ray on the range of 10 nm to 400 nm

The wavelength of photon is given by;

E = hf

f is the frequency of the wave = c / λ

E = h\frac{c}{\lambda}\\\\ \lambda = \frac{hc}{E}

Where;

c is the speed of light = 3 x 10⁸ m/s

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

(a) 0.3 eV = 0.3 x 1.602 x 10⁻¹⁹ J

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(0.3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-6} \ m

λ = 4136 x 10⁻⁹ m

λ = 4136 nm → infrared

(b) 3.0 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-7} \ m

λ = 413.6 x 10⁻⁹ m

λ = 413.6 nm →visible light

(c) 30 eV

\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(30)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-8} \ m

λ = 41.36 x 10⁻⁹ m

λ = 41.36 nm →ultraviolet

5 0
3 years ago
The colors of the stars in the sky range from red to blue. Assuming that the color indicates the frequency at which the star rad
Crazy boy [7]

Answer:

a) 4458K b) 5048K, c) 6166K, d) 5573K

Explanation:

The temperature of the stars and many very hot objects can be estimated using the Wien displacement law

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a) indicate that the wavelength is

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b) lam = 570 nm = 5.70 10⁻⁷ m

    T = 2,898 10⁻³ / 5.70 10⁻⁷

    T = 5084K

c) lam = 470 nm = 4.70 10⁻⁷ m

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    T = 6166K

d) lam = 520 nm = 5.20 10⁻⁷ m

    T = 2,898 10⁻³ / 5.20 10⁻⁷

    T = 5573K

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