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mina [271]
3 years ago
11

A lab technician uses laser light with a wavelength of 650 nmnm to test a diffraction grating. When the grating is 42.0 cmcm fro

m the screen, the first-order maxima appear 6.09 cmcm from the center of the pattern. How many lines per millimeter does this grating have?
Physics
1 answer:
Bingel [31]3 years ago
6 0

Answer:

221 lines per millimetre

Explanation:

We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.

Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,

tanθ = 6.09 cm/42.0 cm = 0.145

From trig ratios, cot²θ + 1 = cosec²θ

cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969

sinθ = 1/cosecθ = 1/6.969 = 0.1435

Also, sinθ = mλ/d at the first-order maximum, m = 1. So

sinθ = (1)λ/d = λ/d

Equating both expressions we have  

0.1435 = λ/d

d = λ/0.1435

Now, λ = 650 nm = 650 × 10⁻⁹ m

d = 650 × 10⁻⁹ m/0.1435

d = 4529.62 × 10⁻⁹ m per line

d = 4.52962 × 10⁻⁶ m per line

d = 0.00452962 × 10⁻³ m per line

d = 0.00452962 mm per line

Since d = width of grating/number of lines of grating

Then number of lines per millimetre = 1/grating spacing

= 1/0.00452962

= 220.77 lines per millimetre

≅ 221 lines per millimetre since we can only have a whole number of lines.

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117,30m

Explanation:

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Solution :

Let us assume that the equilibrium reaction be:

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The expression of K_c will be,

K_c=\frac{[Z][X]^2}{[R][T]^2}

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[R] = concentration of R at equilibrium = 2.0 M

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Answer: 0.85 s

Explanation:

Vs=24.6m/s

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Use equation  for acceleraton to find out time .

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ta=Vf-Vs

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