ray optics. every point on a luminous or illuminated surface produces light rays in all available...

Ray optics

Every point on a luminous or illuminated surface produces light rays in all

available directions

The image in a plane mirror is called a virtual image, meaning that no light rays pass through the image. Rather light rays only appear to originate at the image.

The paraxial approximation: All rays are nearly parallel to the optic axis.

I.e. all angles are small

Spherical mirrors

The small angle approximations are valid whenever the diameter of the mirror is small compared to its radius of curvature. In this case, tan() is approximately equal to , so that

f R / 2

The Principal of Reciprocity:

If a light ray follows a certain path through an optical system, an incident ray in the opposite direction will follow the reversed path.

A ray going through the focal point will emerge parallel to the optic axis.

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Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 40cm from the mirror. Find the location and height of the image.

Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 40cm from the mirror. Find the location and height of the image.

Solution: f = 12cm,

1/si + 1/so = 1/f => si = f so / (so – f)

=> si = (12cm) (40cm) / [ 40cm – 12cm]

= 480 cm2 / 28 cm = 17.1cm

hi = M ho = - (si / so ) ho

= - (17.1 / 40) (8 mm) = - 3.4 mm

Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 18cm from the mirror. Find the location and height of the image.

Solution: f = 12cm,

si = f so / (so – f)

=> si = (12cm) (18cm) / [ 18cm – 12cm]

= 216 cm2 / 6 cm = 36 cm

hi = M ho = - (si / so ) ho

= - (36 / 18) (8 mm) = - 16 mm

Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 6cm from the mirror. Find the location and height of the image.

Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 6cm from the mirror. Find the location and height of the image.

Solution: f = 12cm,

1/si + 1/so = 1/f => si = f so / (so – f)

=> si = (12) (6cm) / [ 6cm – 12cm]

= 48 cm2 / -6 cm = - 8 cm

hi = M ho = - (si / so ) ho

= - (-8 / 6) (8 mm) = + 10.7 mm

Whenever the paraxial approximation is valid,

1 1 1

s s fi o

Mh

h

s

si

o

i

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si > 0 => real image

si < 0 => virtual image

M > 0 => upright image

M < 0 => inverted image M 1 image is enlarged

M 1 image is diminished