Optics Formula

Optics describe the light propagation in terms of light ray. The light ray in geometrical optics is an instrument which is used to approximate models of how a light will propagate. Light rays tend to bend at the interface of two different medium where re\fractive index changes. The geometrical optics gives us the rules for light propagating through optical devices.

The geometrical optics could be made used to explain the geometrical imaging and aberrations.

Lens is one of such optical device with axial symmetry which allows and re\fracts light ray to either converge or diverge the light beam.

The formula for thin lens is as follows:

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

Where, f is the focal length of the thin lens, v is the image distance and u is the object distance.

Magnification of lens is the process by which an object is enlarged in appearance without increasing the size physically.

The formula which helps in getting the thin lens magnification is given by  m = $\frac{h_{i}}{h_{o}}$

$\frac{height of image}{height of object}$

Power of lens (dioptre) = $\frac{1}{f}$ (in meters)

Examples

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Some Examples on Optics Formula are:

Solved Examples

Question 1: For a short sighted person the far point is 60 cm. Calculate the lens power essential to correct the defect ?
Solution:

The power required for a far point of 60 cm is
$left ( \frac{1}{v} right )+left ( \frac{1}{infty } right )=left ( \frac{1}{f_{e}} right )+left ( \frac{1}{f} right )$
$left ( \frac{1}{v} right )+left ( \frac{1}{80} right )=left ( \frac{1}{f_{e}} right )$
By using both equation we get,
$left ( \frac{1}{f} right )$=$-\frac{1}{80}$
Therefore the power ( D ) = $\frac{- 1}{0.8}$ = – 1.25 D
Question 2: A man is suffering from short sight disease is not able to view any objects distinctly beyond anything more than 1.5 m. Find the required power to correct the eye defect?
Solution:

(focal length of eye) $\frac{1}{fe}$ = ( $\frac{1}{v}$ ) + ( $\frac{1}{150}$)

$left ( \frac{1}{v} right )+left ( \frac{1}{infty } right )=left ( \frac{1}{f_{e}} right )+left ( \frac{1}{f} right )$

By substituting both the equation we get
$left ( \frac{1}{f} right )=-\frac{1}{150}$ cm
$left ( \frac{1}{f} right )=-\frac{1}{1.5}$ m
Power ( D ) = – 0.66
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