Question

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground and polished to a concave hemispherical surface with a radius of 4.15 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod. is located on the axis 25.0 cm to the left of the vertex of the concave surface. Find the position of the image of the arrow formed by paraxial rays incident on the concave surface. X cm to the left of the vertex Find the height of the image. mm Is the image erect or inverted? O inverted erect

Answer 1

The thin lens formula is same for both convex lens and concave lens. However, in case of concave lens the focal length becomes negative. We also know that the r= 2f i.e. f = r/2

Now

$\frac{1}{f} = \frac{1}{u}+ \frac{1}{v}$

Substituting r = 2f

$-\frac{2}{4.15} = \frac{1}{25}+ \frac{1}{v}$

$\frac{1}{v} = -\frac{1}{2.075}-\frac{1}{25} = -0.521$

$v = -1.915$

Position of the image = 1.915 cm from the vetrex on the left side

As we know

$\frac{v}{u} = \frac{h_i}{h_o}$

$h_i= \frac{v}{u} \times h_o$

$h_i= \frac{1.915}{25} \times 0.15 = 0.01149 cm$

The height of the image = 0.01149 cm