Tutorial
FE1002 PHYSIC 2 Tutorial 1– [optic] 2007/2008 Semester 2
FE1002 Physic 2 Tutorial 1 – Optic QUESTION 1: light with a wavelength of 700nm is incident on the face of a fused quartz prism (n = 1.46 at 700nm)at an angle of 75˚ (with respect to the normal to the surface). The apex angle of the prism is 60.0˚. Calculate the a) b) c) d) angle of refraction at this first interface angle of incident at this second interface angle of refraction at the second interface and angle between the incident and emerging rays.
1=n2sin 2.
Hint: Use Snell’s law n1sin Solution: a) n1sin
1=n2sin 2.
60
4
b) total angle inside a rectangular = 360˚ 60˚+60˚+ (90˚+41.4˚)+(90˚+ 3)=3 60˚ c) 1.46sin18.6 = 1sin d)
5= 4
75 (75˚2)+( 43) 3 2
QUESTION 2: a) A glass fiber (n=1.50) is submerged in water (n = 1.33). what is the critical angle for light to stay inside the optical fiber? Hint: For light to stay within the optical fibre, it must be incident on the air-glass interface at angle greater than the critical angle. Use Snell’s law n1sin 1=n2 sin 2 with 2 = 90˚. Solution: 1.50sin
2=1.31
b) Determine the maximum angle for which the light rays incident on the end of the pipe shown in the Figure. Are subject to total internal reflection along the walls of the pipe. Assume that the pipe has total internal reflection of 1.36 and the outside medium is water. Hint: Apply Snell’s law to the entrance end of the fibre
.
Done by Foo Boon Kiat (fbk1987@yahoo.com)
Page 1
FE1002 PHYSIC 2 Tutorial 1– [optic] 2007/2008 Semester 2
•
Solution: c i
2
c
1.36 sin
c
= 1 sin 90˚
2
Then using a right angle triangle, 90˚=42.7˚ Then 1.0sin i=1.36sin 2
which is the refraction rays is equal to 180˚-
c-
˚
QUESTION 3: A narrow beam of white light is incident at 25˚ to the normal of heavy flint glass 5.0 cm thick. The indices of refraction of the glass at wavelength of 400nm and 700nm are 1.689 and 1.642, respectively. Find the width of the visible
FE1002 Physic 2 Tutorial 1 – Optic QUESTION 1: light with a wavelength of 700nm is incident on the face of a fused quartz prism (n = 1.46 at 700nm)at an angle of 75˚ (with respect to the normal to the surface). The apex angle of the prism is 60.0˚. Calculate the a) b) c) d) angle of refraction at this first interface angle of incident at this second interface angle of refraction at the second interface and angle between the incident and emerging rays.
1=n2sin 2.
Hint: Use Snell’s law n1sin Solution: a) n1sin
1=n2sin 2.
60
4
b) total angle inside a rectangular = 360˚ 60˚+60˚+ (90˚+41.4˚)+(90˚+ 3)=3 60˚ c) 1.46sin18.6 = 1sin d)
5= 4
75 (75˚2)+( 43) 3 2
QUESTION 2: a) A glass fiber (n=1.50) is submerged in water (n = 1.33). what is the critical angle for light to stay inside the optical fiber? Hint: For light to stay within the optical fibre, it must be incident on the air-glass interface at angle greater than the critical angle. Use Snell’s law n1sin 1=n2 sin 2 with 2 = 90˚. Solution: 1.50sin
2=1.31
b) Determine the maximum angle for which the light rays incident on the end of the pipe shown in the Figure. Are subject to total internal reflection along the walls of the pipe. Assume that the pipe has total internal reflection of 1.36 and the outside medium is water. Hint: Apply Snell’s law to the entrance end of the fibre
.
Done by Foo Boon Kiat (fbk1987@yahoo.com)
Page 1
FE1002 PHYSIC 2 Tutorial 1– [optic] 2007/2008 Semester 2
•
Solution: c i
2
c
1.36 sin
c
= 1 sin 90˚
2
Then using a right angle triangle, 90˚=42.7˚ Then 1.0sin i=1.36sin 2
which is the refraction rays is equal to 180˚-
c-
˚
QUESTION 3: A narrow beam of white light is incident at 25˚ to the normal of heavy flint glass 5.0 cm thick. The indices of refraction of the glass at wavelength of 400nm and 700nm are 1.689 and 1.642, respectively. Find the width of the visible