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ludmilkaskok [199]
3 years ago
6

Light illuminates two closely spaced thin slits and produces an interference pattern on a screen behind. How will the distance b

etween the fringes of the pattern differ for red light and blue light? 1. The same spacing for both 2. Closer for red light 3. Farther apart for red light
Physics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

3. Farther apart for red light

Explanation:

The distance y between fringes is

y = \dfrac{\lambda L }{d},

where \lambda is the wavelength of light, L is the distance to the screen, and d is the slit separation.

Now, the wavelength \lambda_r of the red light is greater than for the wavelength \lambda_b for blue light:

\lambda_r> \lambda_b,

which means

\dfrac{\lambda_r L }{d}>\dfrac{\lambda_b L }{d},

In other words, the distance between the fringes is greater for red light, which from the options is choice 3.

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A. What are the three longest wavelengths for standing waves on a 240-cm-long string that is fixed at both ends?
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Answer:

a) the three longest wavelengths = 4.8m, 2.4m, 1.6m

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Explanation:

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A circuit is constructed with six resistors and two batteries as shown. the battery voltages are v1 = 18 v and v2 = 12 v. the po
VladimirAG [237]

Answer:

V4=9.197v

Explanation:

Given:

V1= 18v ,V2= 12v ,r1=r5=58ohms ,r2=r6=124ohms , r3=47ohms ,r4= 125ohms

V4= I4R4 = V2/(R4 + R5)×R4

V4= 12×125 /(125 + 58)

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5 0
3 years ago
A light wave passes through an aperture (that is, a narrow slit). When it does so, the degree to which the wave spreads out will
crimeas [40]

Explanation:

Single slit diffraction

Diffraction is the phenomenon of spreading out of waves as they pass through an aperture or around objects. Diffraction occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked. in case of  large apertures the wave passes by or through the obstacle without any significant diffraction.

7 0
3 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
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