Distortion, Curvature of Field, Longitudinal Chromatic Aberration for an Object at Finite Distance

Distortion

When the image of the square or rectangular cross wire system is observed through the lens, it doesn’t exactly same as the object. That means distorted image is produced. This kind of defect is called distortion. This defects arises due to the unequal magnification of lens for the light coming from the different object. It is divided into two categories.

• Pin – cushion
• Barrel shaped distortion

If the stops are used before a lens, the resulting image is Barrel – shaped whereas if the stops are used after the lens, then the resulting image is of Pin – cushion type.

Removal:

This kind of defect can be removed by using combination of same kind of lens and stops at suitable position.

Curvature of Field

It is assumed that th e image of the linear extended object should be linear when it is observed through lenses. But very careful study shows that the mage of linear object appears curved when it is observed through lens. This kind of defect is called curvature of field.

In the figure above, the curved image are shown for linear extended object. Here, the focal length is larger around the principle axis and it decreases when moving upwards and downwards.

Removal:

This kind of defect can be removed by using Petzwal’s condition. From Petzwal’s condition, the curvature of final image formed by a systemof thin lens is given by

$$\frac{1}{R}=\sum_n\frac{1}{\mu_n f_n}$$

Here, \(\mu_n\) and \(f_n\) are the refractive index and focal length of the \(n^{th}\) lens respectively.

For final image to be free from curvature of field R = \(\infty\) and hence $$\sum_n\frac{1}{\mu_n f_n}=0$$

For two lens, $$\frac{1}{\mu_1 f_1}+\frac{1}{\mu_2 f_2}=0$$

Since \(\mu_1\) and \(\mu_2\) are always positive, above condition can be obtained if two lens are of opposite type, i.e. concave and convex. So proper combination of two lens satisfying Petzwal’s condition can remove curvature of field.

Longitudinal Chromatic Aberration for an Object at Finite Distance

Let us consider an object O placed in front of a lens L as shown in the figure below. Coloured images are formed by the lens along the principle axis. The image distance for the violet light is nearer to the lens whereas that of for the red is the larger. If a screen is placed in between these two images at the position XY, the image of least chromatic aberration is formed.

Let u be the object distance and \(V_v\) and \(V_r\) be the image distance for the violet and the red colour of light respectively. Suppose that \(f_r\) and \(f_v\) be the focal length of the lens for the red and violet rays of light, then $$\frac{1}{V_v}+\frac{1}{u}=\frac{1}{f_v}\dotsm{1}$$

$$\frac{1}{v_r}+\frac{1}{u}=\frac{1}{f_r}\dotsm(2)$$

Now, subtracting (2) from (1) , we get

$$\frac{1}{v_v}-\frac{1}{v_r}=\frac{1}{f_v}-\frac{1}{f_r}$$

$$or,\;\;\frac{v_r-v_v}{v_v v_r}=\frac{f_r-f_v}{f_v f_r}$$

Taking \(v_v v_r = v^2\) and, \(f_v f_r = f^2\)

$$\frac{v_r-v_v}{v^2}=\frac{f_r-f_v}{f^2}$$

But, \(f_r-f_v = \omega f\)

Now, \(\frac{v_r-v_v}{v^2}=\frac{\omega f}{f^2} = \frac{\omega}{f}\)

$$or,\;\;v_r-v_v=\frac{\omega v^2}{f}\dotsm(3)$$

It is clearly seen in this case that the longitudinal chromatic aberration depends on the distance of the image and hence on the distance of the object from the lens, in addition to, its dependence on the dispersive power and focal length of the lens.

References:

Adhikari, P.B, Daya Nidhi Chhatkuli and Iswar Prasad Koirala. A Textbook of Physics. Vol. II. Kathmandu: Sukunda Pustak Bhawan, 2012.

Jenkins, F.A and H.E White. Fundamental of optics. New York (USA): McGraw-Hill Book Co, 1976.

wood, R.W. Physical Optics. New York (USA): Dover Publication , 1934.