## Section 47.10 Image Formation Bootcamp

### Subsection 47.10.1 Image Formation by Reflection

###### Problem 47.10.1. Images by Two Mirrors with an Acute Angle Between Them.

Follow the link: Checkpoint 47.1.8.

###### Problem 47.10.2. Images of an Object Placed Between Two Parallel Mirrors Facing Each Other.

Follow the link: Checkpoint 47.1.10.

### Subsection 47.10.2 Curved Mirrors

###### Problem 47.10.3. Image of a Point on Axis of a Concave Mirror Within a Focal Point.

Follow the link: Checkpoint 47.2.9.

###### Problem 47.10.4. Image of a Point on Axis of a Concave Mirror Farther than a Focal Point.

Follow the link: Checkpoint 47.2.11.

###### Problem 47.10.5. Image of a Point on Axis of a Concave Mirror Within a Focal Point.

Follow the link: Checkpoint 47.2.13.

###### Problem 47.10.6. Image of a Point on Axis of a Convex Mirror Farther than a Focal Point.

Follow the link: Checkpoint 47.2.15.

### Subsection 47.10.3 Image Formation by Reflection - Algebraic Methods

###### Problem 47.10.7. Real Image in Front of a Concave Mirror.

Follow the link: Checkpoint 47.3.2.

###### Problem 47.10.8. Virtual Image Behind a Concave Mirror.

Follow the link: Checkpoint 47.3.3.

###### Problem 47.10.9. Placing an Object to Obtain Real Image of Given Size from a Concave Mirror.

Follow the link: Checkpoint 47.3.4.

###### Problem 47.10.10. Image by a Convex Mirror.

Follow the link: Checkpoint 47.3.6.

### Subsection 47.10.4 Image Formation by Refraction

###### Problem 47.10.11. Image by Refraction on a Concave Refracting Surface.

Follow the link: Checkpoint 47.4.7.

###### Problem 47.10.12. Image by Refraction from a Convex Surface.

Follow the link: Checkpoint 47.4.8.

### Subsection 47.10.5 Thin Lens Equation

###### Problem 47.10.13. Location and Magnification of Image from a Convex Lens.

Follow the link: Checkpoint 47.7.1.

###### Problem 47.10.14. Location and Magnification of Image from a Concave Lens.

Follow the link: Checkpoint 47.7.2.

### Subsection 47.10.6 Lensmaker's Equation

###### Problem 47.10.15. Carving a Bi-Concave Lens for a Given Focal Length.

Follow the link: Checkpoint 47.8.2.

### Subsection 47.10.7 Image in Two-Lens System

###### Problem 47.10.16. Image by Two Convex Lenses.

Follow the link: Checkpoint 47.9.2.

### Subsection 47.10.8 Miscellaneous

###### Problem 47.10.17. 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 47.10.18. 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).

Work on layer at a time.

\(1.84\text{ cm}\text{.}\)

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 \(\text{O}_1\text{.}\)

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.

From (1), (2) and the figure it is readily seen that

Now, we put in the numbers to obtain, the depth of the final image from the surface of oil.

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.

###### Problem 47.10.19. 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 47.10.20.

Using algebraic method, find the location and magnification of a small bulb sitting \(50\text{ cm}\) from the lens.

Work out image from lens only first and then use that image as object for the mirror.

\(12\text{ cm}\) to the left of the mirror, \(m = \frac{3}{5}\text{.}\)

We first find the intermediate image due to the lens. Let \(q_1\) be the image distance from the lens. We have

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

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

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.

Since \(m\gt 0\) the image will be upright, and, since \(|m| \lt 1\) the image will be smaller than the object.

###### Problem 47.10.21. 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 47.10.22. 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.

Find the image by the lens first and use that as an object for the mirror.

Image 11 cm behind the mirror, \(m = 0.1\text{,}\) \(h_i = 0.2\text{ cm}\text{.}\)

We first find the intermediate image due to lens. Let \(q_1\) be the image distance from the lens. We have

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

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

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 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.

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.

###### Problem 47.10.23. 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 47.10.24. 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, \(R=25\text{ cm}\text{,}\) \(\Delta = 0.3\text{ cm}\text{.}\)

Start with image in \(\text{M}_1\text{.}\)

\(\frac{R\Delta}{ R-2\Delta}\text{,}\) \(0.31 \text{ cm}\)

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 47.10.25. Below, I will work out the distances.

With \(p=d-\Delta\) and \(f=2/R\text{,}\) the image distance \(q_1\) will be

Therefore,

This will be negative. Therefore, we will get object distance to mirror \(\text{M}_2\) by adding its absolute value to \(d\) and using \(d=R/2\text{.}\)

Now, we work out \(q_2\text{.}\)

This gives

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.

Now, with the numerical values given, we get