Question:

A rectangular glass, a convex lens, and a screen are placed as shown in the figure. A 2.0-cm-high object, A, is attached on the rectangular glass. The distances in the figure are: \(a\)=30 cm, \(b\)=20 cm, and \(c\)=40 cm. Under this condition, an image is focused on the screen. If the focal length of the convex lens is 20 cm, what is the index of refraction of the rectangular glass?

Answer:

The object appears to be closer to the lens due to the index of refraction. This is known as “apparent depth.” The actual depth and apparent depth are \(d_{ac}\) and \(d_{ap}\), respectively. We have following relationship:

\[ n = \frac{d_{ac}}{d_{ap}}\]

where \(n\) is the index of refraction. (You can derive this with Snell’s law and some geometric ideas.) Now, we recall the thin lens equation.

\[ \frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f} \]

where \(d_o\), \(d_i\), and \(f\) are the object distance, the image distance, and the focal length. From the second figure, the object distance is \(b+d_{ap}\). The apparent distance is \(d_{ap}=d_{ac}/n\), so we have

\[ d_{ap} = \frac{30}{n} \]

You can plug in the thin lens equation and solve for \(n\).

\[  \frac{1}{b+\frac{30}{n}}+\frac{1}{d_i}=\frac{1}{f} \]

\[  \frac{1}{b+\frac{30}{n}}=\frac{1}{f}-\frac{1}{d_i} \]

\[  \frac{1}{b+\frac{30}{n}}=\frac{d_i-f}{d_if} \]

\[  b+\frac{30}{n}=\frac{d_if}{d_i-f}   \]

\[  \frac{30}{n}=\frac{d_if}{d_i-f} – b  \]

\[  \frac{n}{30}=\frac{1}{\frac{d_if}{d_i-f} – b}  \]

\[  n= \frac{30}{\frac{d_if}{d_i-f} – b}  \]

Therefore, we obtain the index of refraction:

\[ n= 1.5\]