A deviation of 2° is produced in the yellow ray when prism of crown and flint glass are achromatical

Q: What does achromatic combination mean?

It means the combination of prisms produces deviation without dispersion (no splitting of white light into colors).

Q: Why do we subtract the deviations ($\delta_1 - \delta_2$)?

Because the two prisms are placed inverted relative to each other, so their deviations oppose each other.

Q: What is the relation between dispersion and dispersive power?

Dispersive power $\omega$ is defined as the ratio of angular dispersion $\theta$ to the mean deviation $\delta$, so $\theta = \omega \delta$.

Q: Can we use the formula $\delta = (\mu - 1)A$ for any prism?

This formula is an approximation valid only for thin prisms (prisms with a small refracting angle, typically less than $10^\circ$-$15^\circ$).

Q: Why is the yellow ray considered for mean deviation?

The yellow ray is roughly in the middle of the visible spectrum, so its deviation is taken as the mean deviation for the entire spectrum.

Question

A deviation of $2^\circ$ is produced in the yellow ray when prism of crown and flint glass are achromatically combined. Taking dispersive powers of crown and flint glass are $0.02$ and $0.03$ respectively and refractive index for yellow light for these glasses are $1.5$ and $1.6$, respectively. The refracting angles for crown glass prism will be ......$^\circ$ (in degree). (Round off to the nearest integer)

Sigma Percentile · Accepted Answer

### The Magic of Achromatic Combination

Imagine you want to bend a beam of light using a prism, but you don't want it to split into a rainbow of colors. This splitting is called **dispersion**. To achieve deviation without dispersion, we use a clever trick called an **achromatic combination**.

We take two prisms made of different materials—in this case, crown glass and flint glass—and place them inverted relative to each other. The first prism disperses the light, and the second prism recombines it, while still maintaining a net deviation!

### Balancing the Dispersion

For the net dispersion to be zero, the dispersion produced by the crown glass must perfectly cancel out the dispersion from the flint glass. Mathematically, this means:

$$ 	heta_1 = 	heta_2 $$

We know that dispersion $	heta$ is the product of the dispersive power $\omega$ and the mean deviation $\delta$. Therefore, we can write:

$$ \omega_1 \delta_1 = \omega_2 \delta_2 $$

### Calculating the Deviations

The problem states that the net deviation produced by this combination is $2^\circ$. Since the prisms are inverted, their deviations oppose each other. Thus, the net deviation is:

$$ \delta_{	ext{net}} = \delta_1 - \delta_2 = 2^\circ $$

From our dispersion balance equation, we can express $\delta_2$ in terms of $\delta_1$:

$$ \delta_2 = \frac{\omega_1}{\omega_2} \delta_1 $$

Substituting this into the net deviation equation gives:

$$ \delta_1 - \frac{\omega_1}{\omega_2} \delta_1 = 2^\circ $$

Now, we plug in the given dispersive powers ($\omega_1 = 0.02$ and $\omega_2 = 0.03$):

$$ \delta_1 \left(1 - \frac{0.02}{0.03}ight) = 2^\circ $$

$$ \delta_1 \left(1 - \frac{2}{3}ight) = 2^\circ $$

$$ \delta_1 \left(\frac{1}{3}ight) = 2^\circ $$

$$ \delta_1 = 6^\circ $$

### Finding the Prism Angle

We now know that the crown glass prism produces a mean deviation of $6^\circ$. For a thin prism, the deviation is related to the refracting angle $A$ and the refractive index $\mu$ by the formula:

$$ \delta = (\mu - 1)A $$

Applying this to the crown glass prism ($\mu_1 = 1.5$):

$$ 6^\circ = (1.5 - 1)A_1 $$

$$ 6^\circ = 0.5 A_1 $$

$$ A_1 = 12^\circ $$

And there we have it! The refracting angle of the crown glass prism must be **$12^\circ$** to achieve this specific achromatic combination.

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Visualized Solution

$Achromatic Combination$

$Dispersion and Deviation Relation$

$Net Deviation$

$Solving for δ_{1}$

$Calculation of δ_{1}$

$Refracting Angle of Crown Glass$

The Sigma Insight: Refraction and Dispersion through Prism

The Magic of Achromatic Combination

Balancing the Dispersion

Calculating the Deviations

Finding the Prism Angle

Similar Questions

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Achromatic Combination

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dispersion and Deviation Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Deviation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for δ1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation of δ1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Refracting Angle of Crown Glass