Physics, 04.08.2020 18:01 Hfruit

Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.) (a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?

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Answer from: rina62

$\lambda = 1.667 nm$

$\theta = 0.8681^o$

Explanation:

From the question we are told that

The distance of separation is $d = 0.220 \ mm = 0.00022 \ m$

The is distance of the screen from the slit is $D = 2.60 \ m$

The distance between the central bright fringe and either of the adjacent bright $y = 1.97 cm = 1.97 *10^{-2}\ m$

Generally the condition for constructive interference is

$d sin \tha(\theta ) = n \lambda$

From the question we are told that small-angle approximation is valid here.

So $sin (\theta ) = \theta$

=> $d \theta = n \lambda$

=> $\theta = \frac{n * \lambda }{d }$

Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as

$y = D * sin (\theta )$

From the question we are told that small-angle approximation is valid here.

$y = D * \theta$

=> $\theta = \frac{ y}{D}$

$\frac{n * \lambda }{d } = \frac{y}{D}$

$\lambda =\frac{d * y }{n * D}$

substituting values

$\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }$

$\lambda = 1.667 *10^{-6}$

$\lambda = 1.667 nm$

In the b part of the question we are considering the next set of bright fringe so n= 2

Hence

$dsin (\theta ) = n \lambda$

$\theta = sin^{-1}[\frac{ n * \lambda }{d} ]$

$\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ]$

$\theta = 0.8681^o$

Answer from: Quest

freezing

deposition

condensation

explanation;

melting also called fusion is the process where solid becomes liquid, sublimation is the process where solid becomes gaseous without first changing to liquid, deposition involves change from gaseous to solid state, freezing involves change from liquid to solid, boiling or evaporation involves change of liquid to gaseous state, while condensation is the change from gaseous state to liquid state.

endothermic reactions involves absorption of heat from the surroundings or the environment while exothermic reactions involves loss of heat energy to the surroundings.

in this case, sublimation, evaporation and melting are endothermic reactions (gain energy from the environment) while, freezing, deposition and condensation are exothermic (loss of energy to the surroundings) reactions.

Answer from: Quest

Alinear relationship would look like a straight line on the graph. it could have a positive negative or 0 slope. so it could be a line potentially going in any direction but it must be perfectly straight.

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Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.6...

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