Suppose that the central diffraction envelope of a double-slit diffraction pattern contains 15 bright fringes and the first diffraction minima eliminate (are coincident with) bright fringes. how many bright fringes lie between the first and second minima of the diffraction envelope?

6 bright fringes.

Step-by-step explanation:

We know bright interference fringes occur at

d sinθ =mλ where m is an integer.

Here d = 13×a/2(since there are 13 bright fringes ).

Then for first minimum which occurs at angle θ_1 then a sinθ_1 = λ (m=1)

for second minimum which occurs at angle θ_2 then a sinθ_2 = 2λ (m=2)

then we calculate the values of m for which  θ_1 < θ<  θ_2 for  sinθ_1 < sin_θ<  sinθ_2

λ < 2×mλ/13 <2λ

that is 1<2m/13 <2

the satisfied values of m are 7,8,9,10,11,.12. Thus there are 6 bright fringes.

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

hello umm i don't understand your question can you explain it more pls

step-by-step explanation: