Light of wavelength 550 nm illuminates a double slit. The slit width is 24.0 pm

Question

Light of wavelength 550 nm illuminates a double slit. The slit width is 24.0 pm and the separation distance is 0.160 mm. A screen is located 120.0 cm away. Find the theoretical maximum number of interference fringes across the screen. Find the size of the central bright interference fringe on the screen. Find the location of the first diffraction minimum on the screen. Find the relative intensity of the second order bright interference fringe. How many interference fringes are visible in the central bright diffraction region? How many interference fringes are visible in the second order side fringe of the diffraction pattern?

Explanation / Answer

A.}

Maximum path difference in YDSE is equal to separtion between slits

Thus x|max = 0.160 mm = 160×10^3 nm

Thus maximum possible phase difference can be written as k x|max = 2 x|max /lambda

= 2 × ( 160×10^3 /550 ) = 2 × (290.9)

So maximum no. Of bright fringes = (290×2 +1) = 581 ...( 290 up 290 down and 1 central fringe , note that each phase difference of 2 adds two bright fringe to central bright fringe)

No. Of dark fringe = 291×2 = 582 ( between two consecutive bright there is a dark fringe and note that greatest integer of 2× 290.9 is 581 so last fringe will be dark on both side since phase difference comes odd multiple of

So total fringe = 1163

B}.

Size of central fringe will be equal to fringe width = (lambda )× L / slit separation .

Thus ...... Size of central fringe = 550× 10^-9 × 1.2 / 0.00016 m = 4.125 mm

3}

The location of first minium = fringe width / 2 = 2.06mm

4}

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