A 0.50-mm-wide slit is illuminated by light of wavelength 500 nm. What is the width of the central maximum on a screen 2.0 m behind the slit?

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hotcoal3z · Accepted Answer

A single slit width a has bright central maximum of width:  w=2λLα  So:    w=2×500×10−9×20.5×10−3  Result: 4mm

alenahelenash · Accepted Answer

A single slit of width a has a bright central maximum of width: w=2λLa So: w=2×500×10−9×20.5×10−3=4×10−3m Result: 4 mm

user_27qwe · Accepted Answer

Step 1 Given Slit width = 0.50 mm The wavelength of light = 500 nm. Step 2 Here, the width of the central maximum of the single-slit diffraction pattern is given below, w=2λLa Now, w=width of central maximum λ = wavelength of the light a=slit width L=separation between slit and screen Now, convert wavelength from Nano-meter to the meter is shown below, λ=500nm =(500nm)(1.0×10−9m1.0nm) =500×10−9m Step 3 Convert slit width from Mille meter to meter is given below, a=(0.50mm)(1.0m1000mm) =0.0005 m Substitute 0.0005m for a, 500×10−9m for λ and 2.0m for L in w=2λLa. w=2(500×10−9m)(2.0m)0.0005m =4.0×10−3m =(4.0×10−3m)(1mm10−3m) Therefore, the width of the central maximum on screen is 4.0 mm

A 0.50-mm-wide slit is illuminated by light of wavelength 500 nm. What is the wi

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