For nth bright fringe of red light, dsin‌θr=nλr For (n+1) th bright fringe of blue light dsin‌θb=(n+1)λb Since, the fringes coincide, i.e. θr=θb ‌⇒‌‌dsin‌θr=dsin‌θb ‌⇒‌‌nλr=(n+1)λb ‌⇒‌‌n×7.5×10−5=(n+1)5×10−5 ‌⇒‌‌7.5n=5n+5 ‌⇒‌‌2.5n=5⇒n=2