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Korvikt [17]
2 years ago
12

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

wide-that is, the two first-order diffraction minima are separated by 1.40cm
What is the distance between the two second-order minima?
Physics
1 answer:
andrew11 [14]2 years ago
8 0

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1

Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

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the refractive index of cooking oil is 1.47, and the refractive index of water is 1.33. A thick layer of cooking oil is floating
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When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).

The value of this critical angle can be derived by Snell's law, and it is equal to
\theta_C = \arcsin ( \frac{n_2}{n_1} )
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In our problem, n1=1.47 and n2=1.33, so the critical angle is
\theta_C = \arcsin( \frac{1.33}{1.47} )=\arcsin (0.91)=65^{\circ}
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4. How long will it take a car travelling with a speed of 160 km hr to cover a distance of 700 meters? Hint: km/hr should be con
Inessa [10]

Answer:

15.8 seconds

Explanation:

Create an extended calculation to convert all the unit to what you need.

160 km      1000 m       1 hour         1 min

----------- x ------------- x -------------- x ----------   =  44.4 m/s

1 hour            1 km         60 min      60 sec

So 160km/hr is equal to 44.4m/s

Now you can figure out how many seconds it will take to go 700 meters.

44.4 m          

----------   X     x sec   =  700 m

1  sec

Solve for x sec

x sec = 700m / 44.4 m/s

         =  15.8 seconds

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3 years ago
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