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kicyunya [14]
3 years ago
9

What is the thinnest film (but not zero) of MgF2 (n = 1.38) on glass that produces a strong reflection for orange light with a w

avelength of 597 nm ? Express your answer with the appropriate units.
Physics
1 answer:
Art [367]3 years ago
6 0

The concept required to solve this problem is related to thin film interference.

The path length difference between two waves one is reflected from top surface of the film and the bottom surface of the film is equal to twice the thickness of the film:

\gamma = 2t

Where,

\gamma = Path length difference

t = thickness

At the same time we have that the constructive interference condition for a thin film interference for strongly reflected rays is

\gamma = m\lambda

Where

\gamma = Path length difference

\lambda = wavelength

m = Any integer (order of the equation) which represent the number of repetition of the spectrum

Equation both expression we have

2t = m\lambda

Re-arrange to find the thickness we have

t = m\frac{\lambda}{2n}

Our values are given as

m = 1

\lambda= 597nm

n = 1.38

Replacing,

t = (1)\frac{597}{2(1.38)}

t = 216.3nm

Therefore the thinnest thickness of the film is 216.3nm

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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
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Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

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Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

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Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

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v_0 \to LT^{-1} --- dimension

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Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

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a_0 = LT^{-2 --- dimension

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Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

j_0 = LT^{-3} --- dimension

j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

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S_0 = LT^{-4 --- dimension

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Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

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