Physics-
General
Easy

Question

A beam of light has a small wavelength spread delta lambda about a central wavelength lambda. The beam travels in vaccum until it enters a glass plate at an angle theta relative to the normal to the plate, as shown in figure. The index of refraction of the glass is given by n(lambda). The angular spread dq' of the refracted beam is given by

  1. delta theta to the power of ' end exponent equals open vertical bar fraction numerator 1 over denominator n end fraction delta lambda close vertical bar=1nδλ'/>    
  2. delta theta to the power of ' end exponent equals open vertical bar fraction numerator d n invisible function application left parenthesis lambda right parenthesis over denominator d lambda end fraction delta lambda close vertical bar=dn(λ)dλδλ'/>    
  3. delta theta to the power of ' end exponent equals open vertical bar fraction numerator tan invisible function application theta to the power of ' end exponent over denominator n end fraction fraction numerator d n left parenthesis lambda right parenthesis over denominator d lambda end fraction delta lambda close vertical bar=tanθ'ndn(λ)dλδλ'/>    
  4. delta theta to the power of ' end exponent equals open vertical bar fraction numerator sin invisible function application theta over denominator sin invisible function application theta to the power of ' end exponent end fraction fraction numerator delta lambda over denominator lambda end fraction close vertical bar=sinθsinθ'δλλ'/>    

The correct answer is: delta theta to the power of ´ end exponent equals open vertical bar fraction numerator tan invisible function application theta to the power of ´ end exponent over denominator n end fraction fraction numerator d n left parenthesis lambda right parenthesis over denominator d lambda end fraction delta lambda close vertical bar

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