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OPTO 251- MOCK EXAM

1.      There are two good ways to find a description of an image in a lens.  One is to use the thin lens equation. The other is to use

A.     Snell’s law.

B.     The law of refraction.

C.     A free-body diagram.

D.     A ray diagram.

2.      A luminous object that is 4.0-cm high is placed a distance of 45.7 cm from a double convex lens having a focal length of 15.2 cm. Determine the image distance and the image size.

We know that:

h o = 4.0 cm do = 45.7 cm f = 15.2 cm

We need to determine: d I and h i

We can use the lens equation to determine the image distance.

1/f = 1/do + 1/d i

1/ (15.2 cm) = 1/(45.7 cm) + 1/d i

0.0658 cm-1 = 0.0219 cm-1 + 1/d i

0.0439 cm-1 = 1/d i

3.      The visible region of the electromagnetic spectrum extends from about ________ up to about _________

A.     10.0 nm ; 75.0 nm

B.     220 nm ; 530 nm

C.     400 nm ; 700 nm

D.     900 nm ; 12,500 nm

4. The critical angle is the angle of incidence beyond which

A.   All light energy is absorbed.

B.    The frequency of the light approaches zero.

C.   All incident light is reflected.

D.   The medium of incidence becomes transparent.

5. If parallel light rays strike a + 4.00D lens, where will the image be?

Parallel light has no vergence. Therefore, using the equation U + D = V

U = vergence of object at the lens = 0.00D

D = lens power = + 4.00D

Vergence of image rays = V = 0.00D + (+4.00D) = +4.00D.

Converting to centimeters, 100cm/+4.00D = +25 cm to the right of the lens.

7. Find the angle of refraction r for the following diagram.

Snell's law gives:

Rearranging:

Substituting in values:

To get sin 25 degree use the sin button on your calculator.

Here, you use the unfortunately named sin–1 button on your calculator – usually inv and sin buttons.

(We say ‘r is the angle whose sine is 0.282' or ‘r is arcsin 0.282'.)