In Young's double-slit experiment, the ratio of intensities of a bright band and a dark band is 16:1 . The ratio of amplitudes of interfering waves will be

Question

In Young's double-slit experiment, the ratio of intensities of a bright band and a dark band is 16:1 . The ratio of amplitudes of interfering waves will be (A) 16:1 (B) 4:1 (C) 3:1 (D) 5:3

Tardigrade Admin · Accepted Answer

Since (Im a x ./Im i n .)=(16/1) ((. √I1 + √I2 .)2/(. √I1 - √I2 .)2) = (16/1) √I1 + √I2 = 4 √I1 - 4 √I2 A1 + A2 = 4 A 1 - 4 A2 3 A1 = 5 A2 ⇒ (A1/A2) = (5/3)

Q. In Young's double-slit experiment, the ratio of intensities of a bright band and a dark band is $16 : 1$ . The ratio of amplitudes of interfering waves will be

$16 : 1$

$4 : 1$

$3 : 1$

$5 : 3$

Since $\frac{I _{ma x .}}{I _{min .}} = \frac{16}{1}$
$\frac{( I _{1} + I _{2} ) ^{2}}{( I _{1} - I _{2} ) ^{2}} = \frac{16}{1}$
$I_{1} + I_{2} = 4 I_{1} - 4 I_{2}$
$A_{1} + A_{2} = 4 A_{1} - 4 A_{2}$
$3 A_{1} = 5 A_{2}$
$\Rightarrow \frac{A _{1}}{A _{2}} = \frac{5}{3}$

Q. In Young's double-slit experiment, the ratio of intensities of a bright band and a dark band is 16:1 . The ratio of amplitudes of interfering waves will be

16:1

4:1

3:1

5:3