FE1002 PHYSIC 2 Training 1– [optic] 2007/2008 Session 2

FE1002 Physic two Tutorial one particular – Optic QUESTION you: light with a wavelength of 700nm is incident in its appearance of a fused quartz prism (n = 1 . 46 at 700nm)at an angle of 75˚ (with esteem to the usual to the surface). The pinnacle angle in the prism is definitely 60. 0˚. Calculate the a) b) c) d) angle of refraction at this first software angle of incident with this second user interface angle of refraction with the second software and position between the incident and rising rays. 1=n2sin 2 .

Touch: Use Snell's law n1sin Solution: a) n1sin

1=n2sin 2 .

sixty

4

b) total perspective inside a square = 360˚ 60˚+60˚+ (90˚+41. 4˚)+(90˚+ 3)=3 60˚ c) 1 . 46sin18. 6 = 1sin d) 5= some

75 (75˚2)+( 43) three or more 2

PROBLEM 2: a) A glass fiber (n=1. 50) is submerged in water (n = 1 ) 33). precisely what is the essential angle intended for light to settle inside the optic fiber? Sign: For lumination to stay inside the optical dietary fibre, it must be incident on the air-glass interface by angle higher than the important angle. Make use of Snell's legislation n1sin 1=n2 sin 2 with two = 90˚. Solution: 1 ) 50sin 2=1. 31

b) Determine the maximum angle for which the light light incident at the end of the water line shown inside the Figure. Will be subject to total internal reflection along the surfaces of the pipe. Assume that the pipe provides total inside reflection of just one. 36 plus the outside channel is normal water. Hint: Apply Snell's law to the entrance end from the fibre

.

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Webpage 1

FE1002 PHYSIC two Tutorial 1– [optic] 2007/2008 Semester 2

•

Remedy:

c i

2

c

1 . 36 sin

c

= you sin 90˚

2

Then using a proper angle triangle, 90˚=42. 7˚ Then 1 ) 0sin

i=1. 36sin two

which is the refraction sun rays is comparable to 180˚-

c-

˚

QUERY 3: A narrow column of light light can be incident at 25˚ to the normal of heavy flint glass your five. 0 cm thick. The indices of refraction from the glass in wavelength of 400nm and 700nm happen to be 1 . 689 and 1 . 642, correspondingly. Find the width of the visible beam as it comes forth from the piece. Hint: Since n is unique for different wavelength, the sides of refraction for both wavelengths will likely differ. Apply Snell's regulation to equally wavelengths. Answer:

25˚ to a1 a2 5cm 25˚

For beam 1(n=1. 689), the refraction angle is usually 14. 49˚ while the beam 2(n=1. 642) is 16. 92˚ Done by Foo Boon Kiat ([email protected] com) Web page 2

FE1002 PHYSIC two Tutorial 1– [optic] 2007/2008 Semester 2

The difference of distance among ray 1 and beam 2 . Employing right viewpoint triangle.

QUESTION4: A seafood is in a depth g under normal water. Take the index of refraction of water as 4/3. Show that whenever the seafood is viewed at an angle of refraction you, the noticeable depth z . of the. seafood is z= Hint: Using equal aspect triangle, the z percentage of z . and m can w found. X D Ur z z X r 1

m

2

Solution: z = rcos X=rsin

1=Rsin two

1

and d=Rcos

two

Snell's legislation, 1sin sama dengan

1=4/3sin

two

=

QUERY 5: The peak of the true image with a concave mirror is 4 times the thing height if the object is usually 30. 0cm in front the mirror. a)What is the radius of curvature of the reflection? b) Use a ray plan to locate this image. Sign: magnification = image height/ object level. Done by Foo Boon Kiat ([email protected] com) Page 3

FE1002 PHYSIC 2 Guide 1– [optic] 2007/2008 Session 2

Option: a)

reflect equation. After that v=4(30) wherever 30is thing distance.

f=24. and radius of curvature is 2r=48cm. b) target image farrenheit

QUESTION6: A transparent photographic slide is put in front of a converging contact lens with a key length of installment payments on your 44cm. The lens forms an image with the slide 12. 9cm through the slide. How far is the lens from the glide if the picture is a)real b)virtual. Hint: using reflect equation and solve to get the distance thing from the contact lens, p. Solution: for case a ) The image is definitely real which will mean in back of the contact lens, hence. to get case b)the image is within front from the lens, queen is negative.. p = negative worth is declined because p must be before the lens.

ISSUE 6: Two...