\(\frac{I_0}{2}\)
\(\frac{I_0}{4}\)
\(\frac{I_0}{8}\)
Relation between intensities is \(I_R=\big(\frac{I_0}{2}\big)cos^2 (45^\circ)\)
\(=\frac{I_0}{2} \times \frac{1}{2}=\frac{I_0}{4}\)
Therefore, The correct answer is (C) : \(\frac{I_0}{4}\)
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.
Let \( A = \{-3, -2, -1, 0, 1, 2, 3\} \). A relation \( R \) is defined such that \( xRy \) if \( y = \max(x, 1) \). The number of elements required to make it reflexive is \( l \), the number of elements required to make it symmetric is \( m \), and the number of elements in the relation \( R \) is \( n \). Then the value of \( l + m + n \) is equal to:
The molar mass of the water insoluble product formed from the fusion of chromite ore \(FeCr_2\text{O}_4\) with \(Na_2\text{CO}_3\) in presence of \(O_2\) is ....... g mol\(^{-1}\):