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For the grating in Problem 7.13, calculate the wavelengths of first- and second-order diffraction at reflective angles of (a) 25°and (b) 0°. Assume the incident angle is 45°.

Short Answer

Expert verified

a)𝝀1=12.5μm𝝀2=6.27μmb)𝝀1=7.85μm𝝀2=3.92μm

Step by step solution

01

Given information

The given values are

lines/mm=90i=45°a)r=25°b)r=0°

02

To obtain d

d=1mm90linesd=1×10-3m90linesd=1.11×10-5md=11.1×10-6md=11.1μm

03

Part a) Step 1: Calculations for first order diffraction.

Formula

n𝝀=d(sini+sinr)

By putting values in the formula

n=11×𝝀1=11.1μm(sin45°+sin25°)𝝀1=11.1μm(sin45°+sin25°)𝝀1=1.25×10-5𝝀1=12.5μm

04

Part a) Step 2: Calculations for second order diffraction.

Formula

n𝝀=d(sini+sinr)

By putting values

localid="1646387695674" n=22×𝝀2=11.1μm(sin45°+sin25°)𝝀2=11.1μm(sin45°+sin25°)2𝝀2=6.27×10-6𝝀2=6.27μm

05

Part b) Step 1: Calculations for first order diffraction.

Formula

n𝝀=d(sini+sinr)

By putting values

localid="1646387727980" n=11×𝝀1=11.1μm(sin45°+sin0°)𝝀1=11.1μm(sin45°+sin0°)𝝀1=7.85×10-6𝝀1=7.85μm

06

Part b) Step 2: Calculations for second order diffraction.

Formula

n𝝀=d(sini+sinr)

By putting values

localid="1646387762274" n=22×𝝀2=11.1μm(sin45°+sin0°)𝝀2=11.1μm(sin45°+sin0°)2𝝀2=3.92×10-6𝝀2=3.92μm

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