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Shtirlitz [24]
3 years ago
13

Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths f

rom the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.
Physics
1 answer:
Bogdan [553]3 years ago
8 0

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t_{min =  ( m + 1/2) × λ/2n

we substitute

t_{min =  ( 0 + 1/2) × 711 /2(1.21)

t_{min = 0.5 × 711/2.42

t_{min = 0.5 × 293.80165

t_{min = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

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\:  \underline{ \boxed{  { \color{gray}\frak{\huge \: Answer : }}}}

<h3> Collector Current at Saturation :</h3>

  • \sf \large {I_{C(SAT)} = \frac{ V_{CC} }{R_C} }

\:  \:

  • \sf \large {I_{C(SAT)} = \frac{12}{2.2 \times 10 ^{3} } }

\:  \:

  • \bold{ \bf \large {I_{C(SAT)} =5.45mA }}

\:  \:

________________________________

<h3> Value Of Cut - off Voltage : </h3>

  • \sf \large \: V_{CS(cut-off)} = V_{CC}

\:  \:

Therefore ,

\:  \:

  • \bf \large \: V_{CS(cut-off)} = \: 12v

\:  \:

________________________________

{ \bf\large \: Base \:  Current , I_B =  \frac{V_{CC}}{R_C} }

  • \sf \large \: I_B ={ \frac{12}{2.2 \times 10 ^{3} } }

\:  \:

  • \bf \large \: I_B = 50μA

\:  \:

_______________________________

<h3>Collector Current ,</h3>

  • \sf \large \: I_C = β * I_B

\:  \:

  • \sf \large \: I_C = 50 * 50 * 10^{-6}

\:  \:

  • \bf \large \: I_C = 2.5mA

\:  \:

________________________________

<h3> Collector to emitter Voltage : </h3>

  • \sf \large \: V_{CE} = V_{CC} - ( I_C * R_C )

\:  \:  \:

  • \sf \large \: V_{CE} = 12 - ( 2.5 * 10-³ * 2.2 * 10³ )

\:  \:

  • \bf \large \: V_{CE} = 6.5v

________________________________

<h3>Q - point are :</h3>

  • \sf \large \: I_{CEQ} = 2.5mA

\:  \:

  • \sf \large \: V_{CEQ} = 6.5v

________________________________

<h3>Q - point located on the DC load line as shown in fig ~</h3><h3 /><h3 /><h3 /><h3 />

________________________________

Hope Helps!:)

\:  \:  \:  \:

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