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Ira Lisetskai [31]
3 years ago
8

In a bi-prism experiment the eye-piece was placed at a distance 1.5m from the source. The distance between the virtual sources w

as found to be equal to 7.5 x 10-4 m. Find the wavelength of the source of light if the eye-piece has to be moved transversely through a distance of 1.88 cm for 10 fringes.
Physics
1 answer:
Vladimir [108]3 years ago
6 0

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

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Answer:

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7 0
4 years ago
A fixed mass of an ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do
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Answer:

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given data

temperature T1 =  50°C

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solution

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Q  = m ×  c × ΔT     ........................1

here m is mass of the gas and c is specific heat of the gas and Δ T is change in temperature of the gas

here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.

and here at constant pressure Specific heat  is greater than the specific heat at constant volume,

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Specific heat at constant pressure is =  1.005 kJ/kg.K

and

Specific heat at constant volume is =  0.718 kJ/kg.K

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