In Youngs double slit experiment using sodium light (λ =5898 overset texto mathop textA ) , 92 fringes are seen. If given colour (λ =5461 overset texto mathop textA ) is used, how many fringes will be seen:

Question

In Youngs double slit experiment using sodium light (λ =5898 overset texto mathop textA ) , 92 fringes are seen. If given colour (λ =5461 overset texto mathop textA ) is used, how many fringes will be seen: (A) 99 (B) 85 (C) 67 (D) 62

Tardigrade Admin · Accepted Answer

From the relation (n1/n2)=(λ 1/λ 2) (n1/92)=(5898/5461) so, n1=99.36≈ 99

Q. In Youngs double slit experiment using sodium light $(λ = 5898 A o)$ , 92 fringes are seen. If given colour $(λ = 5461 A o)$ is used, how many fringes will be seen:

99

85

67

62

From the relation $\frac{n _{1}}{n _{2}} = \frac{λ _{1}}{λ _{2}}$ $\frac{n _{1}}{92} = \frac{5898}{5461}$ so, $n_{1} = 99.36 \approx 99$

Q. In Youngs double slit experiment using sodium light (λ=5898Ao) , 92 fringes are seen. If given colour (λ=5461Ao) is used, how many fringes will be seen:

99

85

67

62

From the relation n2​n1​​=λ2​λ1​​ 92n1​​=54615898​ so, n1​=99.36≈99

Q. In Youngs double slit experiment using sodium light $(λ = 5898 A o)$ , 92 fringes are seen. If given colour $(λ = 5461 A o)$ is used, how many fringes will be seen:

From the relation $\frac{n _{1}}{n _{2}} = \frac{λ _{1}}{λ _{2}}$ $\frac{n _{1}}{92} = \frac{5898}{5461}$ so, $n_{1} = 99.36 \approx 99$