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

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120 m m

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- 80 m m

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60 m m

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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.

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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 $1.5$ .