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Problem 2: Consider 648 nm light falling on a single slit of width 21 µm. Randomized Variables ? = 648 nm w = 21 µm Part (a) Find the angle, in degrees, of the third diffraction minimum for the light. Numeric: A numeric value is expected and not an expression. ?? = Part (b) What slit width (in micrometers) would place this minimum at 85.0°? Numeric: A numeric value is expected and not an expression. w' =

          Problem 2: Consider 648 nm light falling on a single slit of width 21 µm.
Randomized Variables
? = 648 nm
w = 21 µm
Part (a) Find the angle, in degrees, of the third diffraction minimum for the light.
Numeric: A numeric value is expected and not an expression.
?? = 
Part (b) What slit width (in micrometers) would place this minimum at 85.0°?
Numeric: A numeric value is expected and not an expression.
w' =
        
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Problem 2: Consider 648 nm light falling on a single slit of width 21 µm.
Randomized Variables
? = 648 nm
w = 21 µm
Part (a) Find the angle, in degrees, of the third diffraction minimum for the light.
Numeric: A numeric value is expected and not an expression.
?? = 
Part (b) What slit width (in micrometers) would place this minimum at 85.0°?
Numeric: A numeric value is expected and not an expression.
w' =

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Problem 2: Consider 648 nm light falling on a single slit of width 21 um Randomized Variables 1 = 648 nm w = 21 um Part (a) Find the angle, in degrees, of the third diffraction minimum for the light. Numeric : A numeric value is expected and not an expression. 03= Part (b) What slit width (in micrometers) would place this minimum at 85.0? Numeric : A numeric value is expected and not an expression. W
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please help me with problem, i am not getting the right answer Consider 648 nm light falling on a single slit of width 22 μm.a) Find the angle, in degrees, of the third diffraction minimum for the light. b) What slit width (in micrometers) would place this minimum at 85.0°?

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Transcript

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00:01 In the given problem, we have the value of lambda is equals to, this is the wavelength that is equal to 648 nanometers, or we can say this as 648 multiply 10 to the power minus 9 meter.
00:19 We have the value of width...
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