Consider the interference/diffraction pattern from a double-slit arrangement of slit separation d = 6.60 um and slit width a. Th
e wavelength of the monochromatic light incident normally upon the slits is 2n = d/10 (in air). There is a filter (of negligible thickness) placed on slit 2, so that the magnitude of the EM wave emitted from it is half of that emitted from slit 1. The space between the slits and the screen is filled with water, whose index of refraction is n = 1.33 (you can take noir as 1.00).
(a) What is the wavelength 2 of the light in water?
(b) If a << 1, What is the phase difference between the waves from slits 1 and 2?
(c) For a << 2, Derive an expression for the intensity / as a function of O and other relevant parameters, including the intensity at the center of the screen (where 0 = 0).
(d) Now suppose a = d/3. Redo part (b) above. How many interference maxima are present within the central diffraction peak? (Do not count the "clipped" maxima, if any.) (4) — E -2 d E ) в /Б/ = 1/5 | TT
(a) What is the wavelength 2 of the light in water?
(b) If a << 1, What is the phase difference between the waves from slits 1 and 2?
(c) For a << 2, Derive an expression for the intensity / as a function of O and other relevant parameters, including the intensity at the center of the screen (where 0 = 0).
(d) Now suppose a = d/3. Redo part (b) above. How many interference maxima are present within the central diffraction peak? (Do not count the "clipped" maxima, if any.) (4) — E -2 d E ) в /Б/ = 1/5 | TT
1 answer:
Answer:
(a) λ = 0.496 um (b) S =2π Δ d sinθ/ λ (c) I =gI₀ (d) For the central diffraction peak, a total of 5 interference maxima are present or available.
Note: find an attached copy of a part of the solution to the given question below.
Explanation:
Solution
Recall that:
d = 6.6 um
λ₀ =d/10
λ₀ = 6.6 um
Now,
(a) We find the wavelength λ of the light in water.
Thus,
λ water = (λ₀ )/n
= 0.66/1.33
So,
λ water = λ = 0.496 um
(b) We find the phase difference between the waves from slit 1 and 2
Now,
if a <<d and a<<λ
Then the path difference between the rays will be
Δ S₂N = Δ d sinθ
Thus, the phase difference becomes,
S = 2π Δ/λ is S= 2π Δ d sinθ/ λ
<u>(</u>c) The next step is to derive an expression for the intensity I as function of O and other relevant parameters.
Now,
Let p be the point where these two rays interfere with each other.
Thus,
The electric field vector coming out from slot and and slot 2 is
E₁= E₀₁ cos (ks₁ p - wt) i
E₂ = E₀₂ cos (ks₂ p - wt) i
Note: Kindly find an attached copy of a part of the solution to the given question below.
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Answer:
Explanation:
Given that
Mass of rifle = M
Initial velocity ,u= 0
Mass of bullet = m
velocity of bullet = v
Lets take final speed of the rifle is V
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Initial linear momentum = Final linear momentum
M x 0 + m x 0 = M x V + m v
0 = M x V + m v
Negative sign indicates that ,the recoil velocity will be opposite to the direction of bullet velocity.
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Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
points in the opposite direction of the displacement of the cart, and where
therefore, the work done by the weight is
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Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
therefore, the work done by the weight is
To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.
The relationship that defines both agreements is given by the equation,
Then the frequency for the previous period given (2sec) is
The beat frequency of two frequencies is equal to the difference between the two frequencies, then
<em>Hence option A is incorrect.</em>
We can do this process for 254Hz as and 258 Hz for
, then
<em>Hence option B is incorrect. </em>
We can also do this process for 255Hz as and 257 Hz for
, then
<em>Hence option C is incorrect. </em>
We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then
<em>Hence option D is incorrect. </em>
We can also do this process for 255.75Hz as and 256.25 Hz for
, then
<em>Hence option E is incorrect. </em>
Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz
Answer:
Thus the time taken is calculated as 387.69 years
Solution:
As per the question:
Half life of = 28.5 yrs
Now,
To calculate the time, t in which the 99.99% of the release in the reactor:
By using the formula:
where
N = No. of nuclei left after time t
= No. of nuclei initially started with
(Since, 100% - 99.99% = 0.01%)
Thus
Taking log on both the sides:
t = 387.69 yrs