To solve the problem, we need to find the wavelength of light in a medium with refractive index 1.5 when the light of wavelength 600 nm passes from air into the medium.
1. Relation Between Wavelength and Refractive Index:
The wavelength of light in a medium is given by:
$ \lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{n} $
where $n$ is the refractive index of the medium, $\lambda_{\text{air}}$ is the wavelength in air, and $\lambda_{\text{medium}}$ is the wavelength in the medium.
2. Given Data:
$ \lambda_{\text{air}} = 600\, \text{nm} $
$ n = 1.5 $
3. Calculating Wavelength in Medium:
$ \lambda_{\text{medium}} = \frac{600}{1.5} = 400\, \text{nm} $
Final Answer:
The wavelength of light in the medium is $ {400\, \text{nm}} $.
In a Young’s double slit experiment, a combination of two glass wedges $ A $ and $ B $, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is $ d = 2 \text{ mm} $ and the shortest distance between the slits and the screen is $ D = 2 \text{ m} $. Thickness of the combination of the wedges is $ t = 12 \, \mu\text{m} $. The value of $ l $ as shown in the figure is 1 mm. Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm) with respect to 0 by ____
Two light waves of intensities \(I_1 = 4I\) and \(I_2 = I\) interfere. If the path difference between the waves is 25 % of the wavelength \(\lambda\), find the resultant intensity at that point.