Question

in a young's double slit experiment the wavelength of red light is 7.8*10^-5 cm and that of blue light is 5.2*10^-5 cm. The value of n for which (n+1)th blue bright band coincides with nth bright red band is

Answer 1

Explanation:

Given In a young's double slit experiment the wavelength of red light is 7.8*10^-5 cm and that of blue light is 5.2*10^-5 cm. The value of n for which (n+1)th blue bright band coincides with nth bright red band is

We know that  n λ r =  (n + 1)λb

                So n (7.8 x 10^-5) = n + 1 (5.2 x 10^-5)

                So n + 1 / n = 7.8 x 10^-5 / 5.2 x 10^-5

             Now n + 1 / n = 7.8 / 5.2

                                     = 1.5

                 So n + 1 = 1.5 n

            Now 1.5 n – n = 1

                 0.5 n = 1

              So n = 1/0.5 = 10 / 5

                    So n = 2

Answer 2

Given:

In a young's double slit experiment the wavelength of red light is 7.8*10^(-5) cm and that of blue light is 5.2*10^(-5) cm.

To find:

Value of n for which (n+1)th blue bright band coincides with nth bright red band ?

Calculation:

As per the question, the (n)th red bright fringe will coincide with the (n+1)th blue bright fringe.

$\sf\therefore \: n \times \lambda_{red} = (n + 1) \times \lambda_{blue}$

$\sf\implies \: n \times (7.8 \times {10}^{ - 5} )= (n + 1) \times (5.2 \times {10}^{ - 5} )$

$\sf\implies \: n \times 7.8= (n + 1) \times 5.2$

$\sf\implies \: 2.6(n) = 5.2$

$\sf\implies \: n= 2$

So, value of 'n' is 2 .