CBSE Class 12-science Answered

Young double slit experiment derivation

Asked by Kaushalprince3105 | 16 Feb, 2019, 04:33: PM

Expert Answer

In Young's double slit experiment, two pinholes S1 and S2 (very close to each other) were made on an opaque screen [Fig. (a)].

These were illuminated by another pinholes that was in turn, lit by a bright source. Light waves spread out from S and fall on both S1 and S2.
S1 and S2 then behave like two coherent sources because light waves coming out from S1 and S2 are derived from the
same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S1 and S2.

Thus spherical waves emanating from S1 and S2 will produce interference fringes on the screen GG′, as shown in Fig. (b).

If An arbitrary point P on the line GG′ [Fig. (b)] to correspond to a maximum, then we must have

S₂P – S₁P = nλ; n = 0, 1, 2 ......................... (1)

Now,

begin mathsize 12px style open parentheses S subscript 2 space P close parentheses squared space minus open parentheses S subscript 1 P close parentheses squared space equals space open square brackets D squared plus open parentheses x plus d over 2 close parentheses squared close square brackets space minus space open square brackets D squared plus open parentheses x minus d over 2 close parentheses squared close square brackets space equals space 2 x d space space.................. left parenthesis 2 right parenthesis end style

where S₁S₂ = d and OP = x, then we have

begin mathsize 12px style S subscript 2 P space minus space S subscript 1 P space equals space fraction numerator 2 x d over denominator S subscript 2 P space plus space S subscript 1 P end fraction space space................. left parenthesis 3 right parenthesis end style

If x, d<<D then negligible error will be introduced if S2P + S1P (in the denominator) is replaced by 2D in eqn.(3).

Hence Eq. (3 ) becomes

begin mathsize 12px style S subscript 2 P space minus space S subscript 1 P space equals space fraction numerator x space d over denominator D end fraction end style

........................(4)
Hence we will have constructive interference resulting in a bright region when

begin mathsize 12px style x space equals space x subscript n space equals space fraction numerator n lambda D over denominator d end fraction space semicolon space n space equals space 0 comma space plus-or-minus 1 comma space plus-or-minus 2........ end style

..................................(5)
On the other hand, we will have a dark region when

begin mathsize 12px style x space equals space x subscript n space equals space open parentheses n plus 1 half close parentheses fraction numerator lambda D over denominator d end fraction space semicolon space n space equals space 0 comma space plus-or-minus 1 comma space plus-or-minus 2........ end style

.........................(6)

Thus dark and bright bands appear on the screen. Such bands are called fringes. Equations (5) and (6)
show that dark and bright fringes are equally spaced and the distance between two consecutive bright and dark fringes is given by