Dispersion and Resolving Power of the Prism

4- Dispersion and resolving power of the prism Abstract: We will be conducting an experiment to find the dispersion and resolving power of a prism. We will use a spectrometer lit by an Hg lamp. Light passes through the prism and gets refracted in different angles depending on the wavelength. Introduction: Light can be either monochromatic or polychromatic. Monochromatic light is composed of one wavelength. Upon diffraction, it gets diffracted according to Snell’s law in one direction with no noticeable change in wavelength. Polychromatic light on the other hand is composed of the superposition of multiple waves. When polychromatic light gets refracted, it gets dispersed. We then can see the components of the polychromatic light. This is due to each wave getting refracted in a different angle depending on the wavelength. One of the most widely used optical devices to perform such process is the optical prism. There are many types of optical prisms but one of the most commonly used is the triangular prism, which usually is made of glass and its triangular face is equilateral (𝐴 = 60°). When white light passes through such prism, it gets dispersed into its components, the visible light. The result of the dispersion is known as the spectrum of the polychromatic light. When light passes through a prism, it gets refracted due to the change in velocity twice, once upon entering the prism and once upon exiting. The direction of this bending is determined by the refractive index. The refractive index is a result of the interaction between the electromagnetic wave and the atoms of the material. It is affected by temperature and the wavelength of the incident wave. When white light is incident on the prism, violet waves correspond to a lower refractive index than the red waves. This is why when polychromatic light is incident at the same angle it gets refracted by different angles, making the visible spectrum. The angle of deviation (𝛿) is measured as the angle between the incident ray and the ray exiting the prism after 2 refractions, both following Snell’s law. We can reach the minimum value for 𝛿 when the first refraction results in a symmetrical passing through the prism. we can get: sin * 𝐴+𝛿 𝐴 - = 𝑛 sin * 2 2 𝐴+𝛿 sin / 2 0 𝑛= 𝐴 sin / 2 0 Dispersion: the normal dispersion curve (shown at a previous diagram) has an empirical relation that approximated this curve, it was introduced by Augustin Cauchy: 𝑚 𝑐 𝑛! = 𝑎 + " + # + ⋯ 𝜆 𝜆 Where a, m, c are constants that vary from material to material. Usually, the first two terms are enough to approximate the curve. 𝑛! = 𝑎 + 𝑚 𝜆" Resolving power: The ability of an instrument to form separate images of two neighbouring wavelengths, 𝜆 and 𝜆 + 𝑑𝜆. ℛ= 𝜆 𝑑𝑛 =𝑏 ∆𝜆#$% 𝑑𝜆 Where 𝑏 is the base of the prism. Procedure: 1- Set up the spectrometer as in the diagrams minus the prism. Use an Hg lamp for the light source. 2- Before adding the prism, take the reading of 𝜃$ , which is when you see the white rays using the telescope. Add the prism then. 3- After adding the prism, locate every band of single coloured light using the telescope. Take the reading of 𝜃 corresponding to the colour (wavelength). 4- Graph the relationship between 𝑛 and 𝜆, the slope of the tangent line will help fin the resolving power at that point. % 5- Graph the relationship between 𝑛 and !! . The formula of the best fit line will give us the refractive index 𝑛 as a function of the wavelength 𝜆. Data: 620 1 (× 10%" ) 𝜆" 2.601457 610 2.6874459 38.31 1.5130 594 2.834178 38.4 1.5140 555 3.246490 38.7 1.5174 514 3.785069 39 1.5208 450 4.938272 39.8 1.5298 400 6.250000 40.7 1.5399 𝜆 (nm) 𝐴 = 60° ∆𝜃 ≡ 𝛿 (°) 𝑛 38.27 1.5125 𝑏 = 3.2 𝑐𝑚 The resolving power at 𝜆 = 515: The refractive index: ℛ = 𝑏 𝑠𝑙𝑜𝑝𝑒 = 3232 7.5 × 10&'( 𝑛! = 1.4925 + 𝜆" 𝑛('(%# 7.5 × 10&'( = 1.4925 + = 1.5208 (515 × 10&) )" Sources of error: 1- Accuracy in measurement tools. 2- Human error. Not reading the measurement accurately. 3- Approximation. Error: ∆' (experiment) ' 𝜆 𝑛 (experimental) 𝑛 (equation) ∆𝑛 620 610 594 555 1.5125 1.5130 1.5140 1.5174 1.5120 1.5127 1.5138 1.5168 0.0005 0.0003 0.0002 0.0006 514 450 400 1.5208 1.5298 1.5399 1.5209 1.5295 1.5394 0.0001 0.0066% 0.0003 0.0005 0.0196% 0.0325% 0.0331% 0.0198% 0.0132% 0.0395% Least square method: 𝑥 (× 10%" ) 𝑥𝑦 (× 10%") 𝑦 (n) 𝑥 " (× 10"#) 𝑦 ( (𝑛( ) 2.601457 1.5125 3.934704 6.767579 1.5123 2.6874459 1.5130 4.066106 7.222365 2.834178 1.5140 4.290945 8.032565 1.5130 1.5141 3.246490 1.5174 4.926224 10.5397 1.5172 3.785069 1.5208 5.756333 14.32675 1.5212 4.938272 1.5298 7.554569 24.38653 1.5299 6.250000 1.5399 9.624375 1.5397 ∑ 𝑥 =26.34291 ∑ 𝑦=10.6474 ∑(𝑥𝑦) =40.15326 39.0625 ∑(𝑥 ! ) =110.338 𝑚= 𝑛 ∑(𝑥𝑦) − ∑ 𝑥 ∑ 𝑦 = 7.5119 × 10)%* 𝑛 ∑(𝑥 " ) − (∑ 𝑥 )" 𝑎= ∑𝑦 − 𝑚∑𝑥 = 1.4928 𝑛 𝑦 ( = 7.5119 × 10)%* 𝑥 + 1.4928 7.5119 × 10)%* 𝑛! ′ = + 1.4928 𝜆" Discussion and Conclusion: We were able to find the refractive index as a function over the wavelength. We were also able to find there is resolving power at a certain wavelength. this % was achieved by plotting the relationship between 𝑛 and 𝜆 , 𝑛 and !! . The results we were able to reach were very accurate with a very small margin of error. The results we were able to reach were: 1- The refractive index over material is affected by both the temperature and the wavelength. 2- When light passes through a refractive material, it gets dispersed into its components if it were a polychromatic light because the refractive index for different wavelengths is different. 3- The resolving power of a prism is affected by the wavelength of the light incident on it. 4- We can use Cauchy’s formula to approximate the curve of the relationship between 𝑛 and 𝜆. We can reduce this formula to only the first two terms and still get accurate results. 5- We use a spectrometer to measure the angles between the refracted Rays of different wavelengths. 6- The diffractive index decreases as Lambda increases. % 7- The relationship between the refractive index and !! is linear. 8- The dispersion power decreases as Lambda increases. 9- we can observe dispersion in nature. The formation of rainbows is an example.