6. Matching (10 pts total, 2.5 each) Below are four images of diffraction patter
ID: 2304004 • Letter: 6
Question
6. Matching (10 pts total, 2.5 each) Below are four images of diffraction patterns corresponding diffraction pattern. Write a brief explanation of the defining feature that helped you make the assignment. generated in the lab. For each type of aperture, choose the most likely A. Single Slit Diffraction C. Circular Aperture Diffraction B. Double Slit Diffraction D. Diffraction through a multi- slit diffraction grating 7. (Spts) A laser beam travels into medium 1, refracts into medium 2, undergoes total internal reflection at the boundary of medium 2 and 3 as shown in the picture below. What are the relative sizes of the index of refraction for each medium? TIR n2>n3>n1 B. (5pts) Which lenses below are diverging lenses? Circle all that apply. 1 2 /34Explanation / Answer
6) i) image 1: The pattern formed is circular in nature. This kind of fringe pattern is possible only if the aperture has angular symmetry that is the aperture is circular.(C)
ii)image 2: The pattern formed has maxima and minima just like Double slit interference pattern(Young's double slit experiment) except for the fact that the intensity of maximas is continuosly decreasing. It means that theere is a diffraction envelope over the double slit interfernce pattern. Therefore double slit diffraction pattern.(B)
iii) image 3: This pattern is cleary a single slit differaction pattern as the intensity of second maxima is highly reduced than the first maxima.(A)
iv) image 4: In a multi slit differaction pattern ther are primary maxima and secondry maxima. The central fring is a primary maxima and the other dots shown are secondry maxima. The intensity of secondrys are lower than primarys.
7) When the light ray goes from n1 to n2 it moves away from the normal to the interface. This means that the ray is entering from a denser medium to rarer medium thus n1>n2. TIR can happen only if the light is going from a denser medium to rarer medium n2>n3. therefore n1>n2>n3
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