# Light Questions (Homework 6)

LESSON 6

A ray of light passes through a pane of glass which is 1.0 cm thick. The index of refraction of the glass is 1.53. The angle between the normal to the surface of the pane and the ray in the air as it enters the pane is 67.8°. (a) Find the angle (degrees) between the normal to the surface of the pane and the ray inside the glass. (b) Find the angle (degrees) between the normal to the surface of the plane and the ray in the air after it exits the pane.

If i is the incidence angle and r is the refraction angle into the glass and n is the refraction index then one can write the relation

$sin(i1)/sin(r1) = n$

$sin(r1) = sin(i1)/n$

$sin(r1) = sin(67.8)/1.53 =0.605$

$r =\arcsin(0.605) =37.24 degree$

From the symmetry of the arrangement the refraction angle on the opposite side of the glass (r2) is the same as the incidence angle.

$r2 = 67.8 degree$

A glass fiber ($n = 1.580$) is submerged in water ($n =1.333$). What is the critical angle (degrees) for light to stay inside the optical fiber?

The critical angle is by definition the angle at which

$sin(i) = n$ where I is the incidence angle and n is the refraction index. In this case the relative refraction index is $n = n(glass)-n(water)$

$sin(i) = n(glass) -n(water)$ since the fiber is submerged in water

$i= \arcsin(1.58-1.333) =\arcsin(0.247) =14.3 degree$

An object is placed 13.0 cm in front of a concave mirror with a focal length equal to 3.00 cm. (a) Find the location q (cm) of the image. (b) Find the magnification M. Give the answers to 2 significant figures. Be sure to include the correct signs on theanswers.

For a curve mirror one has the relation

$1/x1+1/x2 = 1/f$

where x1 is the position of the object relative to the mirror, positive is the object is the the left of the mirror (in front of)

$x2$ is the position of the image, positive if the image is to the left of the mirror (in front of)

$f$ is the focal distance of the mirror, positive is the mirror is concave

$1/x2 =1/f-1/x1 =1/3-1/30$

$x2 =3.33 cm$

the magnification of the mirror is by definition (Y1,2 is the length of the object, image)

$M = Y2/Y1 = X2/X1 = 3.33/13 =0.256$ (positive hence the image is non-inverted and smaller)

An object is placed 1.65 cm in front of a concave mirror with a focal length equal to 3.00 cm. (a) Find the location q (cm) of the image. (b) Find the magnification M. Give the answers to 2 significant figures. Be sure to include the correct signs on the answers.

The same considerations as in previous question applies.

$1/x2 = 1/f-1/x1 =1/3 -1/1.65$

$x2 = -3.66 cm$ (negative hence the image is to the right of the mirror (or at the back of or behind the mirror))

$M = X2/X1 =-3.66/1.65 =- 2.22$ (negative hence the image is inverted and larger)

An object is placed 8.0 cm in front of a convex mirror with a focal length equal to −2.0 cm. (a) Find the location q (cm) of the image. (b) Find the magnification M. Be sure to include the correct signs on the answers.

The same consideration as in question 3 applies.

$1/x2 =1/f-1/x1$

$1/x2 =1/(-2) -1/8$

$x2 =-1.6 cm$ (negative hence the image is to the right of the mirror at the back of the mirror)

$M = X2/X1 =-1.6/8 =-0.2$ (the image is inverted and smaller)

An object is placed 3.1 cm in front of a converging lens with a focal length equal to 2.0 cm. (a) Find the location q (cm) of the image. (b) Find the magnification M. Give the answers to 2 significant figures. Be sure to include the correct signs on the answers.

For a lens the following relation applies

$1/x1 +1/x2 =1/f$

where $x1$ is the distance of the object to the lens positive is the object is to the left of the lens (in front of)

$x2$ is the distance of the image to the lens positive is the image is to the right of the lens (behind the lens)

$f$ is the focal distance of the lens positive is the lens is convergent

$1/x2 =1/f -1/x1$

$1/x2 =1/2 -1/3.1$

$x2 = +5.63$ cm (to the right of or behind the lens)

The magnification is by definition

$M = Y2/Y1 = X2/X1 = +5.63/3.1 =+1.81$ (the image is larger and non inverted)

An object is placed 2.3 cm in front of a converging lens with a focal length equal to 6.0 cm. (a) Find the location q (cm) of the image. (b) Find the magnification M. Give the answers to 2 significant figures. The same considerations as in question 6 applies.

$1/x2 =1/f-1/x1$

$1/x2 = 1/6 -1/2.3$

$x2 =-3.73 cm$ (the image is to the left or in front of the lens)

$M = X2/X1 = -3.73/2.3 =-1.62 cm$ (the image is larger and inverted)

An object is placed 9.93 cm in front of a diverging lens with a focal length equal to −5.0 cm. (a) Find the location q (cm) of the image. (b) Find the magnification M. Be sure to include the correct signs on the answers.

The same considerations as in question 6 applies.

$1/x2 = 1/f -1/x1$

$1/x2 =1/(-5) -1/9.93$

$x2 =-3.32 cm$ (the image is to the left of or in front of the lens)

$M = X2/X1 = -3.32/9.93 =-0.33$ (the image is inverted and smaller)

Suppose I want to focus on something 11.0 cm from my eye, but my near point is 25 cm. Find the focal length (cm) of a lens which, when held in front of my eye, will produce an image 25 cm from my eye so that I can focus on it. Neglect the distance between the lens and the eye. Give the answer to 2 significant figures. Be sure to include the correct sign on the answer. The same considerations as in question 6 applies

$1/f = 1/x1+1/x2$

$1/f = 1/11-1/25$

$f = 19.64$ cm positive hence the lens is converging

Find the focal length (cm) of a simple magnifier which has a magnification of 11.0. Use 25 cm for the near point. Assume that the image is at infinity. Give the answer to 2 significant figures.

For a simple Magnifier BY DEFINITION the magnification is

$M = 1/f$

where $f$ is the focal length of the lens

$f = 1/M =1/11 =0.0909 m =9.09 cm$

The focal length of an eyepiece on a microscope is 12 mm. The distance between the eyepiece and the objective lens is 17.1 cm. If the overall magnification of the microscope is −600, find the focal length (mm) of the objective lens. Use 25 cm for the near point. Give the answer to 2 significant figures.

For a microscope BY DEFINITION the magnification is

$M = e/(f1*f2)$

where e is the distance between the focal points of the eyepiece and objective

$f1$ is the focal length of the eyepiece

$f2$ is the focal length of the objective

$e =d -f1-f2$ , where d is the distance between the eyepiece and the objective

$M = (d -f1-f2)/(f1*f2)$

(Attention, here the total magnification is negative!!!!)

$f1*f2*M =d-f1-f2$

$f2(f1*M+1) =d-f1$

$f2 =(d-f1)/(f1*M+1) =(17.1-0.12)/(-0.12*600+1) =$

$=16.98/(-71) =-0.239 cm =-0.24 cm$

Challenge problem (extra credit) A hobby telescope has an objective lens with a focal length of 63 cm and an eyepiece with a focal length of 8.2 mm. We view the planet Jupiter with this telescope. We do this at the time of year when we are closest to Jupiter, a distance of $6.28*10^11$ m. The diameter of Jupiter is $1.40*10^8$ m. (a) Find the diameter (mm) of the image of Jupiter produced by the objective lens. (b) How far (cm) should I place a marble (1-cm diameter) from my eye so that it appears to be the same size as Jupiter as viewed through eyepiece of the telescope? Give the answers to 2 significant figures.
a) for a hobby telescope the grossisment is by definition.

$G = f 2/f1 = 63/0.82 = 76.829$

the magnification
$M = X2/X1 =Y2/Y14 where$x2= 25$cm,$x1 =6.28*10^{11}$m and$Y1 =1.4*10^8$m$M = 0.25/6.28*10^{11} =3.98*10^{-13}$hence the diameter of the image is$Y2 = Y1*M = 1.4*10^8 *3.98*10^{-13} =5.57*10^{-5} m = 5.57*10^{-2} mm$This must be magnified by the grossisment of the hobby telescope$Y2’=G*Y2 =5.57*10^{-2} *76.829 =4.28 mm$b)$x1/x2 = y1/y2x2 =x1*y2/y1 =0.25*0.01/0.00428 =0.5841 m =58.41 cm\$