NEET Wave Optics MCQs & PYQs

The fringe width \( \beta \) in Young's double slit experiment is given by \( \beta = \frac{\lambda D}{d} \). Hence, \( \beta \) increases when slit-to-screen distance \( D \) increases, wavelength \( \lambda \) increases, or slit width separation \( d \) decreases.

The width of the central maximum in single slit diffraction is given by \(w = \frac{2\lambda D}{d}\). Substituting the values: \(w = \frac{2 \times 500 \times 10^{-9} \times 2}{10^{-3}} = 2.0 \times 10^{-3} \text{ m} = 2.0 \text{ mm}\).

The position of \(n^{\text{th}}\) minimum in single-slit diffraction is \(y_n = \frac{n\lambda D}{a}\). The distance between the 1st and 3rd minima is \(Delta y = \frac{2\lambda D}{a} ⇒ a = \frac{2\lambda D}{\Delta y} = \frac{2 \times 6000 \times 10^{-10} \times 0.50}{3 \times 10^{-3}} = 0.2\text{ mm}\).

The fringe width in Young's double slit experiment is given by \( \beta = \frac{\lambda D}{d} \), which is directly proportional to wavelength \( \lambda \). Since \( \lambda_{\text{red}} > \lambda_{\text{green}} > \lambda_{\text{violet}} \), we get \( \beta_1 > \beta_2 > \beta_3 \).

The distance between the first and third minima on the same side of central maximum is \(\Delta y = \frac{2\lambda D}{d}\). Substituting the values: \(3 \times 10^{-3}\text{ m} = \frac{2 \times 6000 \times 10^{-10}\text{ m} \times 0.5\text{ m}}{d}\), we get \(d = 2 \times 10^{-4}\text{ m} = 0.2\text{ mm}\).

The resolving power (RP) of a microscope is given by \(RP = \frac{2NA}{1.22\lambda}\), where \(NA\) is the numerical aperture and \(\lambda\) is the wavelength. Since RP is inversely proportional to \(\lambda\), it will be different for different colours (different wavelengths). Reason (R) states direct proportionality, which is false.

In Young's double slit experiment, intensity is proportional to the square of amplitude \(I \propto A^2\). The amplitude \(A\) is proportional to the slit width \(w\). If \(w_1 = 2w_2\), then \(A_1 = 2A_2\). The ratio of maximum to minimum intensity is \(I_{max}/I_{min} = ((A_1+A_2)/(A_1-A_2))^2\). Substituting \(A_1 = 2A_2\), we get \(I_{max}/I_{min} = ((2A_2+A_2)/(2A_2-A_2))^2 = (3A_2/A_2)^2 = 3^2 = 9\). So, (A) is true. (R) is also true, as amplitude is proportional to slit width. And (R) correctly explains (A).

Assertion (A) describes the Poisson's spot (or Arago spot) phenomenon, a classical example of diffraction where a bright spot appears in the center of the shadow of an opaque circular object. Reason (R) correctly states that this occurs due to constructive interference of light waves diffracting around the edges of the obstacle and meeting in phase at the center of the shadow. Both are true and R explains A.

In YDSE, the intensity of light from a slit is proportional to its width \(I \propto w\). Also, amplitude \(A \propto \sqrt{I}\). If one slit's width increases, its amplitude \(A\) increases. The maximum intensity is \(I_{max} = (A_1+A_2)^2\) and minimum intensity is \(I_{min} = (A_1-A_2)^2\). If \(A_1\) increases, both \(I_{max}\) and \(I_{min}\) will increase. Both A and R are true, and R explains A.

In Young's double slit experiment, the central point has a path difference of zero \(\Delta x = 0\) for all wavelengths \(\lambda\). The condition for constructive interference is \(\Delta x = n\lambda\). For \(n=0\), \(\Delta x = 0\) which is true for all \(\lambda\). Therefore, all colors interfere constructively at the center, resulting in a white fringe. Other fringes are colored due to dispersion. Both are true and R explains A.