In Young's experiment, the fringe width was found to be $0.4 m m$ . If the whole apparatus is immersed in water of refractive index $4 / 3$ , the new fringe width in $m m$ is

0.25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.30

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.40

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.53

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $0.30$
The fringe width in water is
$β_{w} = \frac{β_{a}}{n_{w}}$
$n_{w} =$ refractive index of water $= 4 / 3$
$β_{w} = \frac{0.4}{4 / 3} m m = 0.3 m m$ .

Suggest Corrections

Similar questions

Q. In a Young's double slit experiment,the fringe width is found to be

0.4 m m

. What will be the new fringe width, if the whole apparatus is immersed in a water of refractive index 4/3 without disturbing the geometrical arrangement?

Q. In a Young's double slit experiment, the fringe width is found to be

0.4 m m .

If the whole apparatus is immersed in water of refractive index

\frac{4}{3}

without disturbing the geometrical arrangement, the new fringe width will be

In a Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index $\frac{4}{3}$ , without disturbing the geometrical arrangement, what is the new fringe width?

Q. In a Young's double slit experiment,the fringe width is found to be

0.4 m m

. What will be the new fringe width, if the whole apparatus is immersed in a water of refractive index 4/3 without disturbing the geometrical arrangement?

Q. In a Young’s double slit experiment, the fringe width is found to be 0.4 mm. To make things more interesting, if the whole apparatus is immersed in water of refractive index

\frac{4}{3}

without disturbing the geometrical arrangement, the new fringe width will be:

Join BYJU'S Learning Program

In Young's experiment, the fringe width was found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, the new fringe width in mm is

The correct option is A 0.30The fringe width in water isβw=βanwnw= refractive index of water =4/3βw=0.44/3mm=0.3 mm.

In Young's experiment, the fringe width was found to be $0.4 m m$ . If the whole apparatus is immersed in water of refractive index $4 / 3$ , the new fringe width in $m m$ is

The correct option is A $0.30$
The fringe width in water is
$β_{w} = \frac{β_{a}}{n_{w}}$
$n_{w} =$ refractive index of water $= 4 / 3$
$β_{w} = \frac{0.4}{4 / 3} m m = 0.3 m m$ .