Image Formation by Reflection
1. Images by Two Mirrors with an Acute Angle Between Them.
Follow the link: Exercise 44.1.3.1.
2. Images of an Object Placed Between Two Parallel Mirrors Facing Each Other.
Follow the link: Exercise 44.1.3.2.
Section 44.9 Image Formation Bootcamp
Exercises Exercises
Curved Mirrors
3. Image of a Point on Axis of a Concave Mirror Within a Focal Point.
Follow the link: Exercise 44.2.3.1.
4. Image of a Point on Axis of a Concave Mirror Farther than a Focal Point.
Follow the link: Exercise 44.2.3.2.
5. Image of a Point on Axis of a Concave Mirror Within a Focal Point.
Follow the link: Exercise 44.2.3.3.
6. Image of a Point on Axis of a Convex Mirror Farther than a Focal Point.
Follow the link: Exercise 44.2.3.4.
Image Formation by Reflection - Algebraic Methods
7. Real Image in Front of a Concave Mirror.
Follow the link: Example 44.29.
8. Virtual Image Behind a Concave Mirror.
Follow the link: Example 44.30.
9. Placing an Object to Obtain Real Image of Given Size from a Concave Mirror.
Follow the link: Exercise 44.3.6.1.
10. Image by a Convex Mirror.
Follow the link: Example 44.32.
11. Analyzing Image by a Convex Mirror Algebraically.
Follow the link: Exercise 44.3.6.2.
12. Designing a Magnifying Mirror.
Follow the link: Exercise 44.3.6.3.
13. Deriving the Image-Object-Radius Equation for a Convex Mirror.
Follow the link: Exercise 44.3.6.4.
Image Formation by Refraction
14. Image by Refraction from a Convex Surface.
Follow the link: Example 44.39.
15. Virtual Image by Refraction from a Convex Surface.
Follow the link: Example 44.40.
16. Image by Refraction on a Concave Refracting Surface.
Follow the link: Exercise 44.4.5.1.
Thin Lens Equation
17. Location and Magnification of Image from a Convex Lens.
Follow the link: Example 44.60.
18. Location and Magnification of Image from a Concave Lens.
Follow the link: Example 44.61.
19. Carving a Bi-Concave Lens for a Given Focal Length.
Follow the link: Example 44.62.
20. Image Distance in a Thick Lens.
Follow the link: Example 44.63.
Image in Two-Lens System
21. Image by Two Convex Lenses.
Follow the link: Exercise 44.8.3.1.
22. Virtual Image Through a Two-Convex Lens System.
Follow the link: Exercise 44.8.3.3.
23. A Diverging and Converging Two-Lens System.
Follow the link: Exercise 44.8.3.5.
24. Image formed by a Two-Convex Lens System.
Follow the link: Exercise 44.8.3.4.
Ray Tracing
25. Ray Tracing Through a Convex Lens.
Follow the link: Exercise 44.6.4.1.
26. Ray Tracing Through Two Convex Lenses.
Follow the link: Exercise 44.8.3.6.
27. Ray Tracing Through a Convex Concave Combo.
Follow the link: Exercise 44.8.3.7.
28. Ray Tracing Through a Convex Concave Combo 2.
Follow the link: Exercise 44.8.3.8.
29. Ray Tracing Through a Two Convex Combo.
Follow the link: Exercise 44.8.3.9.
Miscellaneous
30. Apparent Depth when Looking Through Oil and Water Layers.
A small light bulb is at the bottom of a tank that has a layer of water and a layer of oil as shown in Figure 44.73. Let thickess of the oil layer be \(0.4\text{ cm}\) and the depth of bulb from the surface of water be \(2\text{ cm}\text{.}\) Determine the apparent depth of the bulb as seen from near normal. Use a ruler to measure the distance(s).
Hint.
Work on layer at a time.
Answer.
\(1.84\text{ cm}\text{.}\)
Solution.
We can work one interface at a time. Suppose we have only Oil/Water interface. That is the detector is inside oil. Then, we will find that the object will appear at the point \(\text{Q}_1\) with the following relation between the image distance and the object distance from the interface at \(\text{O}_1\text{.}\)
\begin{equation*}
\text{O}_1\text{Q}_1 = \dfrac{n_2}{n_1}\: \text{O}_1\text{P}. \ \ \ (1)
\end{equation*}
Now, we take the image \(\text{Q}_1\) as the object for the interface at \(\text{O}_2\)O. The refraction will now be between the media with refractive indices \(n_2\) and \(n_3\text{.}\) Therefore, the final image at \(\text{Q}_2\) will have the following distance.
\begin{equation*}
\text{O}_2\text{Q}_2 = \dfrac{n_3}{n_2}\: \text{O}_2\text{Q}_1. \ \ \ (2)
\end{equation*}
From (1), (2) and the figure it is readily seen that
\begin{equation*}
\text{O}_2\text{Q}_2 = \dfrac{n_3}{n_2}\: \left[ \text{O}_2\text{O}_1 + \left( \dfrac{n_2}{n_1} \right) \: \text{O}_1\text{P} \right].
\end{equation*}
Now, we put in the numbers to obtain, the depth of the final image from the surface of oil.
\begin{equation*}
\text{O}_2\text{Q}_2 = \frac{1}{1.2} \left( 0.4 + \frac{1.2}{1.33}\times 2 \right) = 1.84\text{ cm}.
\end{equation*}
That is, the bulb appears at only \(1.84\text{ cm}\) from the top of the oil rather than \(2.4\text{ cm}\text{,}\) the actual depth.
31.
Trace rays to find which way the given ray will emerge after refraction through the thin lenses in Figure 44.74. Assume thin lens approximation.
Hint.
Pick a point P on the given ray in each case. Treat that point as an object. Now, find its image Q. Use the rule: All rays on the other side of the lens will either go through Q or appear to be coming from Q.
32.
Copy the figure in Figure 44.75 and draw rays to find the final image.
Hint.
Find the intermediate image through lens alone. Use the intermediate image as the object for the mirror and work with the mirror alone to find the final image.
33. Real Image by a Covex Lens followed by a Cocave Mirror.
A concave mirror of radius of curvature \(10\text{ cm}\) is placed \(30\text{ cm}\) from a thin convex lens of focal length \(15\text{ cm}\) as shown in Figure 44.76.
Using algebraic method, find the location and magnification of a small bulb sitting \(50\text{ cm}\) from the lens.
Hint.
Work out image from lens only first and then use that image as object for the mirror.
Answer.
\(12\text{ cm}\) to the left of the mirror, \(m = \frac{3}{5}\text{.}\)
Solution.
We first find the intermediate image due to the lens. Let \(q_1\) be the image distance from the lens. We have
\begin{equation*}
f_1 = 15\:\text{cm},\ \ p_1 = 50\:\text{cm},\ \ \Longrightarrow\ \ q_1 = \dfrac{150}{7}\:\text{cm} \approx 21.7\text{ cm}.
\end{equation*}
Since \(q_1 \gt 0\) the intermediate image is to the right of the lens and since \(q_1 \lt \text{ separation }\text{,}\) the image is to the left of the vertex of the mirror. Now, this intermediate image will serve as the object for the mirror. By using the separation of the mirror and the lens we find the object distance to the mirror to be
\begin{equation*}
p_2 = 30\:\text{cm} - \dfrac{150}{7}\:\text{cm} = \dfrac{60}{7}\:\text{cm}.
\end{equation*}
The focal length of the mirror is \(f_2 = R/2 = 5\:\text{cm}\text{;}\) it is positive since we have a concave mirror (recall the sign convention). Therefore, the image distance from the mirror will be
\begin{equation*}
\dfrac{1}{q_2} = \dfrac{1}{5}-\dfrac{7}{60},\ \ \Longrightarrow\ \ q_2 = 12\:\text{cm}.
\end{equation*}
Now, sign convenstion for curved mirrors tells us that \(q_2 \gt 0\) for a mirror means that the image is in front of the mirror. Therefore, the final image at \(q_2\text{,}\) will be in front of the mirror at a distance of \(12\text{ cm}\) from the vertex of the mirror.
This image will be real image since it forms in front of the mirror as opposed to image behind the mirror.
The net lateral magnification of the image will come from the product of the magnifications in the lens and the mirror.
\begin{equation*}
m = m_1\times m_2 = \left(- \dfrac{q_1}{p_1} \right)\:\left(- \dfrac{q_2}{p_2} \right) = + \dfrac{3}{5}.
\end{equation*}
Since \(m\gt 0\) the image will be upright, and, since \(|m| \lt 1\) the image will be smaller than the object.
34. Virtual Image by a Cocave Lens followed by a Convex Mirror.
An object of height \(2\text{ cm}\) is placed at \(50\text{ cm}\) in front of a diverging lens of focal length \(40\text{ cm}\) as shown in Figure 44.77. Behind the lens there is a convex mirror of focal length \(15\text{ cm}\) placed \(30\text{ cm}\) from the converging lens. Find the location, orientation and size of the final image.
Hint.
Find the image by the lens first and use that as an object for the mirror.
Answer.
Image \(11\,\text{cm}\) behind the mirror, \(m = 0.1\text{,}\) \(h_i = 0.2\text{ cm}\text{.}\)
Solution.
We first find the intermediate image due to lens. Let \(q_1\) be the image distance from the lens. We have
\begin{equation*}
f_1 = -40\:\text{cm},\ \ p_1 = 50\:\text{cm},\ \ \Longrightarrow\ \ q_1 = -\dfrac{200}{9}\:\text{cm}.
\end{equation*}
Since \(q_1\lt 0\) the intermediate image is to the left of the lens. Now, this intermediate image will serve as the object for the mirror. By using the separation of the mirror and the lens we find the object distance to the mirror to be
\begin{equation*}
p_2 = 30\:\text{cm} - \left( -\dfrac{200}{9}\:\text{cm}\right) = \dfrac{470}{9}\:\text{cm}.
\end{equation*}
The focal length of the mirror is \(f_2 = -15\:\text{cm}\text{,}\) which is negative since the mirror is convex. Therefore, the image distance from the mirror will be
\begin{equation*}
\dfrac{1}{q_2} = \dfrac{1}{-15}-\dfrac{9}{470},\ \ \Longrightarrow\ \ q_2 = -11.7\:\text{cm}.
\end{equation*}
Since \(q\lt 0\) for a mirror means that the image is behind the mirror, the final image at \(q_2\) will be behind the mirror at a distance of \(11.7\,\text{cm}\) from the vertex of the mirror. This image will be a virtual image since it forms behind the mirror where real rays do not go. The net magnification of the image will come from the product of the magnifications in the lens and the mirror.
\begin{equation*}
m = m_1\times m_2 = \left(- \dfrac{q_1}{p_1} \right)\:\left(- \dfrac{q_2}{p_2} \right) = +0.1.
\end{equation*}
Since \(m>0\) the image will be upright and since \(|m|\lt 1\) the image will be smaller than the object. The height of the image is given by the product of \(m\) and the height of the object.
\begin{equation*}
h_i = m\:h_o = (0.1)(2\:\text{cm}) = 0.2\:\text{cm}.
\end{equation*}
35. Image Formation by a Convex LEns and a Concave Mirror.
An object of height \(3\,\text{cm}\) is placed at \(25\,\text{cm}\) in front of a converging lens of focal length \(20\,\text{cm}\text{.}\) Behind the lens there is a concave mirror of focal length \(20\,\text{cm}\text{.}\) The distance between the lens and the mirror is \(5\,\text{cm}\text{.}\) Find the location, orientation and size of the final image.
Solution.
Let us draw the physical layout to help visualize the problem.
Now, we will work out the image \(q_1\) from the lens L which will serve as the object \(p_2\) for the mirror. The image \(q_2\) will be the final image we seek.
From the given data for the object and the lens we get
\begin{equation*}
\dfrac{1}{q_1} = \dfrac{1}{20} - \dfrac{1}{25} = \dfrac{1}{100},\ \ q_1 = +100\:\textrm{cm}.
\end{equation*}
We will subtract \(q_1\) from the separation between the lens and the mirror to find the object distance \(p_2\) for the mirror.
\begin{equation*}
p_2 = 5\:\textrm{cm} - 100\:\textrm{cm} = -95\:\textrm{cm}.
\end{equation*}
Now using the mirror equation for the mirror we get
\begin{equation*}
\dfrac{1}{q_2} = \dfrac{1}{20} - \dfrac{1}{-95} = \dfrac{23}{380},\ \ q_2 = \dfrac{380}{23}\:\textrm{cm} = +16.5\:\textrm{cm}.
\end{equation*}
Since \(q_2 \lt 0\) for a mirror means the image is in front of the mirror. The final image will form in front of the mirror at a distance \(16.5\,\text{cm}\) from the mirror. The image will form as a result of real rays crossing there and therefore will be a real image. The magnification of the final image with respect to the original object will be
\begin{equation*}
m = m_1\: m_2 = \left(- \dfrac{q_1}{p_1}\right)\: \left(- \dfrac{q_2}{p_2}\right) = -0.7.
\end{equation*}
Since \(m \lt 0\) the image would be inverted. Multiplying the height of the object by \(m\) we find that the image has a height of \(2.1\, \text{cm}\text{.}\)
36. Image by Two Converging Lenses and a Concave Mirror.
An object of height \(3\,\text{cm}\) is placed at \(25\,\text{cm}\) in front of a converging lens of focal length \(20\,\text{cm}\text{.}\) Behind the lens there is another converging lens of focal length \(20\,\text{cm}\) placed \(10\,\text{cm}\) from the first lens. There is a concave mirror of focal length \(15\,\text{cm}\) placed \(50\,\text{cm}\) from the second lens. Find the location, orientation and size of the final image.
Answer.
\(27\, \text{cm}\) in front of the mirror, \(m = 0.6\text{,}\) \(h_i= 1.76\, \text{cm}\text{.}\)
Solution.
Let us draw the physical layout to help visualize the problem.
Now, we will work out the image \(q_1\) from lens L1 which will serve as the object \(p_2\) for the lens L2. The image \(q_2\) of the lens L2 will serve as the object \(p_3\) for the mirror. The image \(q_3\) will be the final image we seek.
From the given data for the object and lens L1 we get
\begin{equation*}
\dfrac{1}{q_1} = \dfrac{1}{20} - \dfrac{1}{25} = \dfrac{1}{100},\ \ q_1 = +100\:\textrm{cm}.
\end{equation*}
We will subtract \(q_1\) from the separation between the two lenses to find the object distance \(p_2\) for the lens L1.
\begin{equation*}
p_2 = 10\:\textrm{cm} - 100\:\textrm{cm} = - 90\:\textrm{cm}.
\end{equation*}
Note that the object distance for lens L2 is negative. This occurred due to the object (which is the intermediate image from lens L1) being on the right side of lens L1. Now using the lens equation for L2 we get
\begin{equation*}
\dfrac{1}{q_2} = \dfrac{1}{20} - \dfrac{1}{-90} = \dfrac{11}{180},\ \ q_2 = +16.4\:\textrm{cm}.
\end{equation*}
We will subtract this distance from the distance between the lens L2 and the mirror M to obtain the object distance \(p_3\) for M.
\begin{equation*}
p_3 = 50\:\textrm{cm} - 16.4\:\textrm{cm} = 33.6\:\textrm{cm}.
\end{equation*}
Now, using the mirror equation we find
\begin{equation*}
\dfrac{1}{q_3} = \dfrac{1}{15} - \dfrac{1}{33.6},\ \ q_3 = +27\:\textrm{cm}.
\end{equation*}
Since \(q_3 \gt 0\) for a mirror means the image is in front of the mirror, the final image will form in front of the mirror at a distance \(27\,\text{cm}\) from the mirror. The image will form as a result of real rays crossing there and therefore will be a real image. The magnification of the final image with respect to the original object will be
\begin{equation*}
m = m_1\: m_2\: m_3 = \left(- \dfrac{q_1}{p_1}\right)\: \left(- \dfrac{q_2}{p_2}\right)\: \left(- \dfrac{q_3}{p_3}\right) = 0.59.
\end{equation*}
Since \(m \gt 0\) the image is upright, i.e. it has the same vertical orientation as the object. Multiplying the height of the object by \(m\) we find that the image has a height of \(1.76\,\text{cm}\text{.}\)
37. Real Floating Image of a Penny Placed Between Two Concave Mirrors.
Two identical concave mirrors are placed facing each other with a distance of \(d=R/2\) between their vertices as shown in Figure 44.80. One of them has a small hole in the middle. When a penny is placed on the bottom mirror, you observe a real image of the penny floating above the hole.
Explain how this happens by showing images in the two mirrors and locate the distance above the mirror with the hole that you will see the floating image. If you like, you can use the following numerical values, \(R=25\text{ cm}\text{,}\) \(\Delta = 0.3\text{ cm}\text{.}\)
Hint.
Start with image in \(\text{M}_1\text{.}\)
Answer.
\(\frac{R\Delta}{ R-2\Delta}\text{,}\) \(0.31 \text{ cm}\)
Solution.
Since the penny is located within a focal distance of mirror \(\text{M}_1\text{,}\) a virtual image will be formed at \(q_1\text{,}\) which will serve as an object for mirror \(\text{M}_2\text{.}\) Now, the object distance \(p_2\) from mirror \(\text{M}_2\) is larger than the focal length, hence image will be a real image at \(q_2\text{.}\) This is illustrated in Figure 44.81. Below, I will work out the distances.
With \(p=d-\Delta\) and \(f=2/R\text{,}\) the image distance \(q_1\) will be
\begin{equation*}
\frac{1}{q_1} = \frac{2}{R} - \frac{1}{d-\Delta}.
\end{equation*}
Therefore,
\begin{equation*}
q_1 = \frac{R\,(d-\Delta)}{2 d-2 \Delta - R} = - \frac{R\,(d-\Delta)}{2\Delta}.
\end{equation*}
This will be negative. Therefore, we will get object distance to mirror \(\text{M}_2\) by adding its absolute value to \(d\) and using \(d=R/2\text{.}\)
\begin{equation*}
p_2 = |q_1| + d = \frac{R(d-\Delta)}{2\Delta} + d = \frac{R^2}{4\Delta}.
\end{equation*}
Now, we work out \(q_2\text{.}\)
\begin{equation*}
\frac{1}{q_2} = \frac{2}{R} - \frac{4\Delta}{R^2}.
\end{equation*}
This gives
\begin{equation*}
q_2 = \frac{R^2}{2R-4\Delta}.
\end{equation*}
Now, this is a little bit beyond mirror \(\text{M}_1\text{.}\) We can get the distance above the hole by subtracting \(d\) from it.
\begin{equation*}
\text{above hole} = q_2 - d = \frac{R^2}{2R-4\Delta} - \frac{R}{2} = \frac{R\Delta}{ R-2\Delta}.
\end{equation*}
Now, with the numerical values given, we get
\begin{equation*}
\text{above hole} = 0.31 \text{ cm}.
\end{equation*}
38. Image Through a Convex Lens and a Plane Mirror.
A lamp of height \(5\,\text{cm}\) is placed \(40\,\text{cm}\) in front of a converging lens of focal length \(20\,\text{cm}\text{.}\) There is a plane mirror \(15\,\text{cm}\) behind the lens. Where would you find the image when you look in the mirror?
Solution.
Let us find the image by the lens first. Let \(q_1\) be the image distance of this intermediate image from the lens.
\begin{equation*}
\dfrac{1}{q_1} = \dfrac{1}{f} - \dfrac{1}{p_1},\ \ \Longrightarrow\ \ q_1 = 40\:\textrm{cm}.
\end{equation*}
This image would have formed at a point behind the mirror had the mirror not been in the way. Now, this image can be used as an object for the mirror equation with \(f=\infty\) for the plane mirror. The object distance \(p_2\) for the mirror is now
\begin{equation*}
p_2 = d - q_1 = 15\:\textrm{cm}-40\:\textrm{cm} = -25\:\textrm{cm}.
\end{equation*}
This gives the following for the image distance \(q_2\text{.}\)
\begin{equation*}
q_2 = - p_2 = 25\:\textrm{cm}.
\end{equation*}
The positive \(q\) for a mirror means that the image is formed in front of the mirror rather than behind the mirror. The distance to the image is \(25\,\text{cm}\) from the mirror, or \(10\,\text{cm}\) on the left of the lens.
39. Parallel Rays Striking a Convex Lens at an Angle.
Parallel rays from a far away source strike a converging lens of focal length \(20\,\text{cm}\) at an angle of \(15^\circ\) with the horizontal direction. Find the vertical position of the real image observed on a screen in the focal plane.
Answer.
\(5.4\, \text{cm}\) from the axis.
Solution.
From the triangle \(\triangle\)IOF in the figure it is clear that
\begin{equation*}
\textrm{IF} = \textrm{OF}\:\tan\:\angle\textrm{IOF}
\end{equation*}
Therefore,
\begin{equation*}
\textrm{IF}= (20\:\textrm{cm})\:\tan\:15^{\circ} = 5.4\:\textrm{cm}
\end{equation*}
40. Parallel Rays Striking a Concave Lens at an Angle.
Parallel rays from a far away source strike a diverging lens of focal length \(20\,\text{cm}\) at an angle of \(10^\circ\) with the horizontal direction. As you look through the lens, where in the vertical plane the image would appear?
Answer.
\(3.5\, \text{cm}\) from the axis.
Solution.
From the triangle \(\triangle\)IOF in the figure it is clear that
\begin{equation*}
\textrm{IF} = \textrm{OF}\:\tan\:\angle\textrm{IOF}
\end{equation*}
Therefore,
\begin{equation*}
\textrm{IF}= (20\:\textrm{cm})\:\tan\:10^{\circ} = 3.5\:\textrm{cm}
\end{equation*}
41. Images of an Object Between a Plane Mirror and a Convex Mirror.
A light bulb is placed \(10\,\text{cm}\) from a plane mirror, which faces a convex mirror of radius of curvature \(8\,\text{cm}\text{.}\) The plane mirror is located at a distance \(30\,\text{cm}\) from the vertex of the convex mirror. Find the location of two images in the convex mirror. Are there other images? If so, where are they located?
Solution.
One way to locate images in the convex mirror is to start with the image in the plane mirror and find all the images of images associated with this image as shown in the figure.
Other way is to start with the first image formed in the convex mirror and work out the images of images.
