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lianna [129]
3 years ago
7

A thin film of oil of thickness t is floating on water. The oil has index of refraction no = 1.4. There is air above the oil. Wh

en viewed from the air in the direction normal to the surface, there is constructive interference of reflected red light with l = 675 nm. Find the minimum thickness t of the oil film that will give constructive interference of reflected light with l = 675 nm.
Physics
1 answer:
kkurt [141]3 years ago
7 0

Answer:

t = 120.5 nm

Explanation:

given,    

refractive index of the oil = 1.4

wavelength of the red light = 675 nm

minimum thickness of film = ?

formula used for the constructive interference

2 n t = (m+\dfrac{1}{2})\lambda

where n is the refractive index of oil

t is thickness of film

for minimum thickness

m = 0

2 \times 1.4 \times t = (0+\dfrac{1}{2})\times 675

t = \dfrac{0.5\times 675}{2\times 1.4}

        t = 120.5 nm

hence, the thickness of the oil is t = 120.5 nm

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single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev
Goryan [66]

Answer:

The sound level of the 26 geese is  Z_{26}= 96.15 dB

Explanation:

From the question we are told that

    The  sound level is Z_1 =  81.0 \ dB

     The number of geese is N = 26

Generally the intensity level of sound is mathematically represented as

        The intensity of sound level in dB  for one  goose is mathematically represented as

                       Z_1 = 10 log [\frac{I}{I_O} ]

Where I_o is the  threshold level of intensity with value  I_o = 1*10^{-12} \  W/m^2

            I is the intensity for one goose in W/m^2

For 26 geese the intensity would be  

          I_{26} = 26 * I

   Then  the intensity of 26 geese in dB is  

              Z_{26} = 10 log[\frac{26 I }{I_o} ]

               Z_{26} = 10 log (\ \ 26 *  [\frac{ I }{I_o} ]\ \ )

               Z_{26} = 10 log (\ \ 26  \ \ ) *   (\ \  10 log [\frac{ I }{I_o} ]\ \ )

 From the law of logarithm we have that

              Z_{26} = 10 log 26 +  10 log [\frac{I}{I_0} ]

                    = 14.15 + 82

                    Z_{26}= 96.15 dB

               

               

           

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3 years ago
In the combo circuit diagrammed, R1 = 19.2 Ω, R2 = 20.7 Ω, and R3 = 25.8 Ω. Find the equivalent resistance of the circuit.
garik1379 [7]

Answer:

Equivalent resistance: 13.589 Ω

Explanation:

R series = R1 + R2 + R3 ...

\frac{1}{R_{eq} } = \frac{1}{R1} +\frac{1}{R2} +\frac{1}{R3} ...

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\frac{1}{R_{eq} } = \frac{1}{19.2} +\frac{1}{46.5}\\\\\frac{1}{R_{eq} } = 0.0735887097\\\\R_{eq} = 13.5890411

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