Lenses, mirrors, a glass slab and a prism

Aim

To find the focal length of a convex lens and of a concave mirror from the $u$ and $v$ readings and their graphs. To find the refractive index of a glass slab and of a prism.

The focal length of a convex lens

A convex lens makes a real image of a bright object on a screen. For an object at the distance $u$ and an image at the distance $v$, both measured from the lens, the lens formula gives

Here $u$ and $v$ are written as positive distances, for a real object and a real image. In the New Cartesian sign convention that NCERT uses, $u$ is negative and the formula is $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$. They are the same law. A-Level and IB books use "real is positive", which matches the form above.

The apparatus. An optical bench with a lighted object (a wire gauze or a pin), the lens in a holder, and a screen. All the centres are at the same height on the same line.

The steps.

1. For a rough value, form the image of a distant window on the screen. The distance of the screen is about $f$.
2. Put the object at the distance $u$, say $20$ cm, and move the screen until the image is sharp. Read $u$ and $v$ from the scale. A parallax test with a pin helps.
3. Repeat for $u$ about $25, 30, 35, 40, 50$ cm. The object must be farther than $f$.
4. Note the index correction, the difference between the scale reading and the true distance, from the thickness of the holders. Apply it to $u$ and $v$.

The readings. A lens of about $15$ cm.

The mean is $f = 15.00$ cm.

Two graphs. A graph of $v$ against $u$ is a curve, a hyperbola. It passes through the point $u = v = 2f$. Read $2f$ from the point where the curve meets the line $v = u$, and halve it. A graph of $\dfrac{1}{v}$ against $\dfrac{1}{u}$ is a straight line of slope $-1$, since $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$. It cuts both axes at $\dfrac{1}{f}$. The intercept gives $f = 15.01$ cm.

Move the object from $20$ cm to $60$ cm. What happens to the dot? Where is it when the object is at $2f$?

The displacement method. Fix the object and the screen at a distance $D$ apart, with $D > 4f$. There are two positions of the lens, a distance $d$ apart, that give a sharp image. Then $f = \dfrac{D^2 - d^2}{4D}$. This needs no index correction.

The focal length of a concave mirror

A concave mirror forms a real image of a pin placed beyond its focus. Use a pin $O$ as an object and a second pin $I$ as an image, and remove the parallax between the image and the pin $I$. The formula is the same, $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$, with distances from the pole. When the pin is at the centre of curvature, the image is at the same place and $u = v = 2f = R$.

For a mirror of $f = 12.0$ cm, five readings are

The mean is $f = 12.0$ cm. The same graphs, and the same precautions, apply.

The refractive index of a glass slab

Light passes through a rectangular glass slab of thickness $t$. It bends at the first face, bends back at the second face, and emerges parallel to its first path. The angle of incidence $i$ and the angle of refraction $r$ follow Snell's law:

The steps. Draw a normal on a sheet on a drawing board. Fix the slab and two pins $P_1$ and $P_2$ on the incident ray at an angle $i$. Look through the other face and fix two more pins $P_3$ and $P_4$ in line with the images of $P_1$ and $P_2$. Join $P_3P_4$. Draw the path inside the slab, and measure $r$. Repeat for $i = 20^\circ, 30^\circ, \ldots, 70^\circ$.

A graph of $\sin i$ against $\sin r$ is a straight line through the origin, and its slope is $n = 1.52$. The lateral shift $d$ between the incident and the emergent rays is $d = \dfrac{t\sin(i - r)}{\cos r}$. For $t = 2.5$ cm, $i = 50^\circ$ and $r = 30.5^\circ$, it is $d = 0.97$ cm. A thicker slab or a larger angle of incidence shifts the ray more.

The real and apparent depth. For an object under a slab, $n = \dfrac{\text{real depth}}{\text{apparent depth}}$ near the normal. A travelling microscope measures both.

The refractive index of a prism

Light is bent twice in a triangular prism, with the angle of the prism $A$. The angle of deviation $\delta$ is the angle between the incident ray and the emergent ray. It changes with the angle of incidence $i$, and has a lowest value, the angle of least deviation $\delta_m$. This happens when the ray inside the prism is parallel to the base, so that the angle of incidence equals the angle of emergence, $i = e$. Then $r_1 = r_2 = \dfrac{A}{2}$, and

The steps. Measure $A$ with two pins and a protractor, or with a spectrometer. Fix the pins $P_1$, $P_2$ for each angle of incidence $i$ and $P_3$, $P_4$ for the emergent ray, and measure $\delta$. Draw the graph of $\delta$ against $i$. Its lowest point is $\delta_m$. For two values of $i$, the same $\delta$ occurs, since $i$ and $e$ can swap.

For a prism of $A = 60^\circ$:

The lowest point is $\delta_m = 38.9^\circ$, near $i = 49^\circ$. So $n = \dfrac{\sin 49.45^\circ}{\sin 30^\circ} = 1.52$. The speed of light in the glass is $\dfrac{c}{n} = \dfrac{3.00 \times 10^{8}\ \text{m/s}}{1.52}$ $= 1.97 \times 10^{8}\ \text{m/s}$.

Raise the angle of the prism from $50^\circ$ to $70^\circ$ and watch the lowest point. Then raise the index. Which change makes the deviation larger?

Precautions. Fix the pins vertical and far apart, at least $10$ cm. Look at the bases of the pins. Measure angles with a protractor of least count $0.5^\circ$. Draw the lines with a sharp pencil.

Practise with new numbers

Try it in Python

The first program finds $f$ for a lens from each reading, and from the straight line of $1/v$ against $1/u$.

The second program finds the index of the slab from the line of $\sin i$ against $\sin r$, and the index of the prism from the least deviation.

Common mistakes

• Mixing the signs of $u$ and $v$. In $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ the object distance is negative. In $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$ both distances are taken as positive.
• Taking the sharpest image too quickly. Move the screen to and fro, and choose the middle of the range.
• Leaving out the index correction. The scale readings include the thickness of the holders.
• Using the angles and not their sines. Snell's law uses $\sin i$ and $\sin r$.
• Taking the smallest $\delta$ in the table as $\delta_m$. Draw the graph.
• Using a thin prism formula for a thick prism. $n = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$ is exact at the least deviation.

Remember these

• $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$. The graph of $\dfrac{1}{v}$ against $\dfrac{1}{u}$ is a line that cuts both axes at $\dfrac{1}{f}$. At $u = v = 2f$ the image is as large as the object.
• Displacement method: $f = \dfrac{D^2 - d^2}{4D}$ for $D > 4f$.
• Slab: $n = \dfrac{\sin i}{\sin r}$. Lateral shift $d = \dfrac{t\sin(i - r)}{\cos r}$.
• Prism: $n = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$, with $i = e$ at the least deviation.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A convex lens forms a real image at $60$ cm from the lens for an object at $20$ cm. What is its focal length?

2. A ray enters a glass slab at $45^\circ$ and is refracted at $28^\circ$. What is the refractive index of the glass?

3. For a prism, the angle of incidence at the least deviation is $48^\circ$ and the least deviation is $36^\circ$. What is the angle of the prism?

4. Why do we take the focal length from a graph and not only from one pair of $u$ and $v$?

References

• NCERT, Physics Class XII, Chapter 9 (Ray Optics and Optical Instruments), the sections on the lens formula, refraction and the prism.
• NCERT, Laboratory Manual: Physics, Class XII, the experiments on the focal length of a lens and of a mirror, the refractive index of a glass slab and of a prism.
• OpenStax University Physics Volume 3, Chapter 2 (Geometric Optics and Image Formation), the sections on thin lenses and mirrors.