ray optics & optical instruments

light

• electromagnetic radiations (400-750 nm)

• travels in a straight line

• speed in air/ vacuum = c = 3 × 10^8 m/s

laws of reflection of light

• angle of incidence = angle of reflection (angle between the rays and the normal)

• the incident ray, reflected ray and normal all lie on the same plane

sign convention for concave lens

• u = -

• v = + (virtual), - (real)

• f = -

• hi = + (virtual), - (real)

• ho = +

• m = + (virtual), - (real)

sign convention for convex lens

• u = -

• v = + (always virtual)

• f = +

• hi = + (always virtual)

• ho = +

• m = + (always virtual)

proof that f = R/2

derive it

mirror formula

• 1/f = 1/v + 1/u

v = uf/ u-f

mirror equation derivation (using concave mirror)

derive it

mirror equation derivation (with convex mirror)

derive it

magnification derivation

derive it

vector form of reflected ray

r^ = i^ - 2(i^. n^).n^

r = reflected ray, i = incident ray, n = normal

minimum length of mirror to see your image

mirror length = H/2

(mirror has to be H/2 above the ground)

angle of deviation (\delta )

• 1 mirror: \delta = \pi - 2i

  • (anti-clockwise rotation)

• 2 mirrors: \delta = 2\pi - 2\theta

how mirror rotation affects the rotation of reflected ray

if mirror rotates by \theta , the reflected ray rotates by 2\theta

image velocity (plane mirror)

assuming velocities of image, object and mirror to be perpendicular to the mirror,

Vi = 2Vm - Vo

no. of images formed by 2 mirrors

• n = 360/\theta (if 360/\theta is odd)

• n = 360/\theta - 1 (if 360/\theta is even)

\theta = angle between both mirrors

diameter of distant objects (like sun)

d = \thetaf (\theta in radians)

image velocity (spherical mirror)

assuming Vm = 0,

Vi = -(f²/u - f)Vo

calculating distances measured from focus

f² = x1x2

(x1, x2 are distances of object and image from f)

transverse magnification

m = hi/ ho = -v/u = -f/ u-f

longitudinal magnification

m = image length along principal axis/ obj. length along principal axis

what happens to the image when part of a spherical mirror is covered?

• taking the laws of reflection to be true for all points on the remaining parts of the mirror, the image will still be that of the full object

• however, the intensity would be reduced as the area of the surface has been reduced (if half the mirror has been covered, the intensity becomes half)