Ray Optics and Optical Test - 6

A telescope will and can not change the size of an object big or small. What it do is to magnify them if viewed from far. It is like zooming a photo. So to an observer the tree would appear 10 times closer. In fact the image in the eye is 10 times taller through the telescope than through the naked eye. That makes it seem 10 times closer.

Magnification of telescope can be increased by using eyepiece of lower focal length (f_e). Since power is inversely proportional to focal length, eyepiece of large power will increase magnification of the telescope.

Resolving power is directly proportinal to aperture.

Image will be formed at the same position and same height but intensity of image formed will be less hence its brightness will be less as less number of light rays will form the image. Light rays from the covered portion will not contribute to image formation.

Let h be the actual height and h' be the apparent height of bird at any instant.
Then, h/h′ = μ_aw (refractive index of air with respect to water) = 3/4 (since refractive index of water with respect to air is 4/3)
If v is the actual speed and v' be the apparent speed of bird, then
v = dh/dt and v' = dh'/dt
v/v' = (dh/dt) / (dh'/dt) = 3/4
or v' = 4v/3

It is like a combination of two Plano – convex lenses. Therefore only one image is formed.

When light refracts through a lens with a spherical surface, the rays that pass through the outer diameters of the lens will focus at a different point from rays that pass through the center. This causes the resulting image to be blurred. Circular annular mask blocks out rays that blur the image.

The rays from the point object fall normally on the surface of the sphere and hence emerge undeflected. When produced backward, they meet at O. Thus a virtual image is formed at center itself.

Therefore, distance of virtual image from the surface of the sphere = 6cm.