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

Question

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

Tardigrade Admin · Accepted Answer

As, (I max /I min )=((a+ b)2/(a-b)2)=(16/1) ∴ (a +b/a-b)=(4/1) 4 a-4 b=a+ b or 3 a=5 b ∴ (a/b)=(5/3)

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

16

5/3

4

1/4

As, $\frac{I _{m a x}}{I _{m i n}} = \frac{( a + b ) ^{2}}{( a - b ) ^{2}} = \frac{16}{1}$
$∴ \frac{a + b}{a - b} = \frac{4}{1}$
$4 a - 4 b = a + b$
or $3 a = 5 b ∴ \frac{a}{b} = \frac{5}{3}$

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

16

5/3

4

1/4

As, Imin​Imax​​=(a−b)2(a+b)2​=116​ ∴a−ba+b​=14​ 4a−4b=a+b or 3a=5b∴ba​=35​

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

As, $\frac{I _{m a x}}{I _{m i n}} = \frac{( a + b ) ^{2}}{( a - b ) ^{2}} = \frac{16}{1}$
$∴ \frac{a + b}{a - b} = \frac{4}{1}$
$4 a - 4 b = a + b$
or $3 a = 5 b ∴ \frac{a}{b} = \frac{5}{3}$