1, On Earth, flint glass has an n = 1.73. A sample (thickness: 1.4 mm) of that s
ID: 1787192 • Letter: 1
Question
1, On Earth, flint glass has an n = 1.73. A sample (thickness: 1.4 mm) of that same glass made in space refracts a light beam 1.8° less than one made on Earth. Given that the incident beam is 30° for both: Determine: a. b. c. the angle of refraction in space the total distance traveled by the transmitted beam within the glass the velocity of the refracted beam 2. A blue-green argon laser light is being evaluated. The first maximum occurs at a distance of 5.26 mm on a screen 2.34 m away using a diffraction grating with 4500 slits/m. Determine: a. b. c. the wavelength of the light the angular displacement the angular displacement of the 3rd maximumExplanation / Answer
1.
======================================
a. let r be the refracted ray on earth
then r on space = ( r-1.8)
using snells law n1 sin i = n2 sin r
sin r = 1 * sin 30/ 1.73
sin r = 0.289
r = 16.8 deg
then refractive index on space = 16.8 -1.8 = 15 deg
------------------- ---------------------
b. Possible distance travelled inside is 1.4 mm
------------------------------------
c.use n = c/V
v = c/n
v =(3*10^8)/(1.73)
v = 1.73 *10^8 m/s
=============================================
2.
a. use Y = m LR/d
L is wavelength
R is distance from slit to screen = 2.34 m
d is slit width = 1/N = 1/4500 = 2.22*10^-4 m
Y = fringe width = 5.26 mm
so L = Y d/mR
L = (5.26 *10^-3 * 2.22 *10^-4)/(1 * 2.34)
L = 500nm
---------------------------
b. use d sin theta = mL
sin theta = 1 * 500*10^-9/(2.22*10^-4)
angnular displacement theta = 2.25*10^-3 rad
-----------------------------
c.
for m= 3,
sin theta = 3* 500 *10^-9/(2.22*10^-4)
theta = 6.75 *10^-3 radians
=====================================================
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.