For possible interference maxima on the screen, the condition is $ d\sin \theta =n\lambda $ ... (i) Given: $ d=slit-width=2\lambda $ $ \therefore $ $ 2\lambda \sin \theta =n\lambda $ $ \Rightarrow $ $ 2\sin \theta =n $ The maximum value of $ \sin \theta $ is $ 1 $ , hence, $ n=2\times 1=2 $ Thus, E (i) must be satisfied by 5 integer values $ ie, $ $ -2,\,\,-1,\,\,0,\,\,1,\,\,2 $ . Hence, the maximum number of possible interference maxima is $ 5 $ .