One end of a cylindrical glass rod ends in a hemispherical convex surface of radius $R = 20 m m$ . Find the image distance of a point object on the axis of the rod, placed $80 m m$ outside its vertex. The rod is in the air and the refractive index of glass is $1.5$ .

120 m m

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

120 m m

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

- 80 m m

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

60 m m

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

Open in App

Solution

The correct option is A $120 m m$
Given:
$μ_{1} = 1$
$μ_{2} = 1.5$
$R = + 20 m m$
$u = - 80 m m$

Using spherical surface formula,
$\frac{1.5}{v} - \frac{1}{- 80} = \frac{1.5 - 1}{+ 20}$
$v = + 120 m m$

The image is therefore formed at the right of the vertex (v is positive) and at a distance of 120 mm from it.

Suggest Corrections

Similar questions

One end of a cylindrical glass rod ends in a hemispherical surface of radius R = 20 mm.
- Find the image distance of a point object on the axis of the rod, 80 mm to the left of the vertex. The rod is in air.

Q. One end of a cylindrical glass rod ends in a hemispherical surface of radius

R = 20 m m

. Find the image distance (in mm) of a point object on the axis of the rod,

80 m m

to the left of the vertex. The rod is in air.

Q. One end of a cylindrical glass rod ends in a hemispherical surface of radius R = 20 mm. The rod is immersed in water of index 1.33. Find the image distance if object distance is 80 mm.

Q. A hemispherical surface of radius

R = 20 m m

. Find the image distance (in mm) of a point object on the axis of the rod,

80 m m

to the left of the vertex. The rod is in air.

Q. One end of a cylindrical rod is grounded to a hemispherical surface of radius

R = 20 m m .

It is immersed in water of refractive index

1.33.

If the refractive index of the hemisphere is

1.5.

Find the position of an image of an object placed on the axis of the rod inside water at

10 c m

from the pole.

Join BYJU'S Learning Program

One end of a cylindrical glass rod ends in a hemispherical convex surface of radius R=20 mm. Find the image distance of a point object on the axis of the rod, placed 80 mm outside its vertex. The rod is in the air and the refractive index of glass is 1.5.

The correct option is A 120 mmGiven: μ1=1 μ2=1.5 R=+20 mm u=−80 mm Using spherical surface formula, 1.5v−1−80=1.5−1+20 v=+120 mm The image is therefore formed at the right of the vertex (v is positive) and at a distance of 120 mm from it.

One end of a cylindrical glass rod ends in a hemispherical convex surface of radius $R = 20 m m$ . Find the image distance of a point object on the axis of the rod, placed $80 m m$ outside its vertex. The rod is in the air and the refractive index of glass is $1.5$ .