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Delicious77 [7]
3 years ago
14

Monochromatic light of wavelength 550 no passes through a diffraction grating that has 55000 lines/m. A bright line appears on t

he screen at an angle of 12.23o from the central bright fringe. What is the order of this bright line?
A. 5
B. 6
C. 7
D. 8
Physics
1 answer:
just olya [345]3 years ago
7 0

Answer:

Hence the order of the bright fringe is 7.

Explanation:

Diffraction  can  be defined as the deviation from rectilinear propagation and  diffraction  grating is used to measuring the wavelength of light using a spectrometer.

We have the formula  d sin∅ = n l

We have to find the value of n in this equation and all other values are given.

Here d = 55000 lines

∅ = 12.23°

Wavelength , l = 550 nanometer

n is the order of the bright line.

Substitute these values in the given equation,

55000\times sin(12.23°) = n\times 550 \times 10^{-9}

n = 7

Hence the order of the bright fringe is 7.

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Answer:

A.  xmax = 131.49 m

B.  t = 8.74 s

C.  ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

y=y_o+v_osin\theta-\frac{1}{2}gt^2      (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2 (2)

You use the quadratic formula to obtain the value of t:

t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

x_{max}=v_ocos\theta t      

x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}     (3)

By replacing in the equation (3) the values of all parameters you obtain:

y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m

The maximum height reached by the cannon ball is 220.33m

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