An optical two-slit experiment is setup with its slits separated by 48 X 10-3 cm

Question

An optical two-slit experiment is setup with its slits separated by 48 X 10-3 cm and produces a second order maximum at an angle of 0.09o.

a) What is the wavelength of light used in this experiment? 377Nm OK

b) What would the wavelength of the light need to be if we kept the position of the second order maximum the same, but increased the slit separation of 69 X 10-3 cm?
542Nm it says no. Why???? What did i do wrong?

**c)What would the angle to the second maximum be if we used the wavelength calculated in part (b) but kept the original slit separation?** I don't know this one either.

Appreciate any help thanks!!!!

Explanation / Answer

GIVEN Seperation of slits ( d ) = 48 X 10-3 cm Position of second order maxima = 0.09o ........................................................................................ (a) Bright fringe condition in Young's douuble slit experiment sin = m ( / d ) For second order maxima m= 2 sin (0.09o ) = 2 ( / 48 X 10-3 cm )                = 377 nm --------------------------------------------------------------------------- (b) If d = 69 X 10-3 cm sin = m ( / d ) sin (0.09o ) = 2 ( / 68 X 10-3 cm )             = 534 nm ----------------------------------------------------------------------- (c)  As per given problem If   = 534 nm d = 48 X 10-3 cm Then sin () = 2 ( 543 nm / 48 X 10-3 cm )                         =22.62 x 10-4                 = sin-1 (22.62 x 10-4 )                   = 0.129o ........................................................................................ (a) Bright fringe condition in Young's douuble slit experiment sin = m ( / d ) For second order maxima m= 2 sin (0.09o ) = 2 ( / 48 X 10-3 cm )                = 377 nm --------------------------------------------------------------------------- (b) If d = 69 X 10-3 cm sin = m ( / d ) sin = m ( / d ) sin (0.09o ) = 2 ( / 68 X 10-3 cm )             = 534 nm ----------------------------------------------------------------------- (c)  As per given problem If   = 534 nm d = 48 X 10-3 cm Then sin () = 2 ( 543 nm / 48 X 10-3 cm )                         =22.62 x 10-4                 = sin-1 (22.62 x 10-4 )                   = 0.129o                 = sin-1 (22.62 x 10-4 )                   = 0.129o

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