If the 5 text th order maxima of wavelength 400 nm in YDSE coincides with n text th order maxima of the wavelength 5000 mathringA, then the value of n is

Question

If the 5 text th order maxima of wavelength 400 nm in YDSE coincides with n text th order maxima of the wavelength 5000 mathringA, then the value of n is (A) 3 (B) 4 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

(5 λ1 D/d)=(n λ2 D/d) 5 × 4000=n × 5000 n=4

Q. If the $5^{th}$ order maxima of wavelength $400 nm$ in YDSE coincides with $n^{th}$ order maxima of the wavelength $5000 \overset{˚}{A}$ , then the value of $n$ is

3

4

5

6

$\frac{5 λ _{1} D}{d} = \frac{n λ _{2} D}{d}$
$5 \times 4000 = n \times 5000$
$n = 4$

Q. If the 5th order maxima of wavelength 400nm in YDSE coincides with nth order maxima of the wavelength 5000A˚, then the value of n is

3

4

5

6

d5λ1​D​=dnλ2​D​ 5×4000=n×5000 n=4

Q. If the $5^{th}$ order maxima of wavelength $400 nm$ in YDSE coincides with $n^{th}$ order maxima of the wavelength $5000 \overset{˚}{A}$ , then the value of $n$ is

$\frac{5 λ _{1} D}{d} = \frac{n λ _{2} D}{d}$
$5 \times 4000 = n \times 5000$
$n = 4$