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Given that, slit = 0.01mm Angle of incidence (beta) = 20 degree lambda = 532 nm m = 1 For 0 degree incidence angle the condition for interference is d sin theta = m lambda It the incidence angle Ts then, d sin theta = m lambda + d sin Ts for m = 1, Ts = 20 degree, d = 0.01 mm, lambda = 532 nm sin theta_1 = 532 times 10^-9/1 times 10^-7 + 1 times 10^-3/1 times 10^-3 sin 20 = 0.395 theta_1 = sin 6-1 (0.395) = 23.3 degree therefore theta_1 = 23.5 degree This may helps you.
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