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Lelu [443]
3 years ago
15

Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and th

e distance between constructive fringes on a screen 5 m away is 0.3 cm. What is the wavelength?
Physics
1 answer:
Alik [6]3 years ago
8 0

Answer:

The wavelength of the light is 7200\ \AA.

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes \beta= 0.3\ cm

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

\beta=\dfrac{D\lambda}{d}

Put the value into the formula

0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}

\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}

\lambda=7.2\times10^{-7}\ m

\lambda=7200\ \AA

Hence, The wavelength of the light is 7200\ \AA.

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Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0
3 years ago
? Which statement is true of an object in equilibrium?
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7 0
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Experiment with the battery voltage set to 15 volts, measure the current in a parallel circuit with 1,2,3, and 4 light bulbs. (I
koban [17]

Answer:

1) current = I

2) Resistance = V/I

3) current = 2I

4) resistance = V/2I

5) current = 3I

6) Resistance = V/3I

7) Current = 4I

8) Resistance = V/4I

Explanation:

When one bulb is connected across the battery then let say the current is given as I

Then resistance is given as

R = \frac{V}{I}

When two bulbs are in parallel with the battery then

total current becomes twice of initial current

so we have

current = 2I

Resistance of the circuit is now

R = \frac{V}{2I}

When three bulbs are in parallel with the battery then

total current becomes three times of initial current

so we have

current = 3I

Resistance of the circuit is now

R = \frac{V}{3I}

When four bulbs are in parallel with the battery then

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so we have

current = 4I

Resistance of the circuit is now

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3 0
3 years ago
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