Physics Homework Assignment: Lens, Mirrors, and Diffraction Analysis

This document presents a comprehensive solution to a physics assignment, addressing key concepts in optics, lenses, mirrors, and diffraction. The solution begins with calculations involving the lens formula and magnification, followed by an analysis of angular radius and the Rayleigh criterion. It explores the properties of converging lenses and spherical mirrors, including their applications and image formation characteristics. The assignment also delves into the principles of diffraction, specifically examining slit spacing and the wavelengths of diffracted light using a diffraction grating. Furthermore, the solution addresses the concept of f-number and depth of field in photography, comparing different scenarios and their effects on image focus. The solution concludes with references to relevant physics principles and equations.

Question 1:
Figure 1: FBD Sketch
Figure 2: Lens Sketch
D
2 πd = A
360 , whereA= Angular
h
2 π ×17 = A
360 ,henceA = h
17 × 360
2 π
From the given data,
f = 8cm
u = 7cm
From lens formula,
1
v + 1
u = 1
f
OR

1
v = 1
f −1
u
Substituting the values of f and u:
1
v = 1
8 − 1
7 =−1
56
Hence, v = -56 cm
Magnification:
M =−v
u = h '
h =−−56
7 = h '
h = 8
1 =8 ,where:
h = height of the object, and
h’ = height of the image
h '
2 πd = AL
360 ° ,
d = (v+10) = 66 cm
Hence, h'
2 π ×66 = AL
360 ° => AL= h '
66 × 360 °
2 π

Question 2
0.01= θ−sinθ
θ
1− sinθ
θ =0.01
sinθ
θ =0.99
Now for
θ=0.245 rad
sinθ
θ =0.99
Hence
θ=0.245 rad
Question 3
(a) Using small angle approximation,
The moon is located at =384400km from the earth. Its radius is 3474/2=1737. Its
angular radius from the earth =1737/384400=0.004519
Because 1 radian =57.3 degrees, the angular radius of the moon
=0.004519*57.3=0.2589degress.
0.2589degress=15.534 minutes of an arc
(b) From Rayleigh Criterion
θ=1.22 λ
b = 1.22× 500 ×10−9
25.08× 103 =2.4322 ×10−13
Since the darkness-adapted pupil is 7mm, the resolution is adequate to see the crater.
(c) The lens has a larger aperture. The larger the aperture of a lens the better the
resolution of the lens. It will thus be possible to view the crater using the lens.

Question 4
(a) Converging lenses are thicker at the centers compared to their edges. They are
thus used to converge light at the focal point behind the lens. Their workings
are based on light refraction principles where light is bent twice. Spherical
lenses have the inner and outer surfaces that are spherical in shape.
Because the converging lens’s focal length is 55mm, the film has to be placed 55mm
from the lens in order to obtain a clear and a sharp image for an objective that is
placed at infinity, a property of lens where the rays from an object become parallel to
the principle axis and then meet at the focus resulting into a small, diminished and an
inverted object.
(b) From the mirror formula, we have
1
f = 1
d0
+ 1
di
Therefore,
1
0.055 = 1
1.2 + 1
di
hence 1
di
=17.33
di=0.0576 mtowardsthemirror
(c) The spherical mirror should be made such that the outer edge of parabolic mirror
have a significantly different shape than that of the spherical mirror. This will help in
creating a sharp and clear images that are devoid of blurriness.

Question 5
(a)
f
22 = 1
30
f
x = 1
25 0
From which x, the number becomes
x= 250 ×22
30 =183.33
f/number must be f/183.33 in order to maintain the same amount of light into the film
in the second film because the apertare has to be maintained to get the same amount
of fight.
b. The other difference will be on the depth of the field (Dof) refers to how much
content in a picture appears to be in focus.
In the first image (f/22, 1/30 sec)because of a relatively smaller apertaure most of the
picture will be in focus that is (the main item will be in focus while the background
will be less blurred). This forms a good condition for taking photos of stationary
objects that have stationary background.
In the second picture, a wider aperture would result in shallow DOF that limits the
focus area (confined to the object) while most of the background is blurred. The first
the main item will be in focus while the background will be less blurred

Question 6
(a) dsinθ=nλ
For the first order, n = 1
Slit spacing of differential grating d=0.002× 10−3 m
The first line is observed at the angle:
θ1=17 ° 7.7 minofarc=17+ 7.7
60 =17.1283 °
Wavelength of the first line is:
λ1=dsinθ1=0.002 ×10−3 sin ( 17.1283° ) =589.0259× 10−9 m=589.0259 nm
This corresponds to yellow color(Chapter 6 -page 1, n.d.).
(b) First order, n = 1
Wavelength of the second line is:
λ2=589.5924 nm
θ2= λ2
dsin
Hence,
θ2=sin−1
( λ2
d ) =sin−1
( 589.5924 ×10−9
0.002 ×10−3 )=17.1453 °=17 ° 8.72 minofarc
(c) For second order, n = 2.
θ1=sin−1
( n λ1
d )=sin−1
( 2× 589.0259× 10−9
0.002 ×10−3 )=36.088 °
θ2=sin−1
( n λ2
d )=sin−1
( 2 ×589.5924 ×10−9
0.002 ×10−3 )=36.128 °
The same pair of lines are again observed for the second spectrum (n = 2)
References
Chapter 6 -page 1. (n.d.). Retrieved from
http://pages.cs.wisc.edu/~dyer/ah336/papers/06_color.pdf