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 the air.

120 mm

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-120 mm

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-80 mm

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60 mm

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Solution

The correct option is A 120 mm

$μ_{1} = 1, μ_{2} = 1.5, R = + 20 m m, u = - 80 m m$
$\frac{1.5}{v} - \frac{1}{- 80} = \frac{1.5 - 1}{20}$
$v^{'} = + 120 m m$
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 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 the air.

The correct option is A 120 mm μ1=1,μ2=1.5,R=+20 mm,u=−80 mm 1.5v−1−80=1.5−120 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.

The correct option is A 120 mm

$μ_{1} = 1, μ_{2} = 1.5, R = + 20 m m, u = - 80 m m$
$\frac{1.5}{v} - \frac{1}{- 80} = \frac{1.5 - 1}{20}$
$v^{'} = + 120 m m$
The image is therefore formed at the right of the vertex (v is positive) and at a distance of 120 mm from it.