In Young’s double slit interference pattern, the fringe width increases with
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.
A beam of light of wavelength \(\lambda = 500 \text{ nm}\) from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The width of central maxima is
The width of the central maximum in single slit diffraction is given by \(w = \frac{2\lambda D}{d}\). Substituting the values: \(w = \frac{2 \times 500 \times 10^{-9} \times 2}{10^{-3}} = 2.0 \times 10^{-3} \text{ m} = 2.0 \text{ mm}\).
A screen is placed 50 cm from a single slit, which is illuminated with 6000 Å light. If distance between the first and third minima in the diffraction patten is 3 mm, the width of the slit is
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 \(Delta y = \frac{2\lambda D}{a} ⇒ a = \frac{2\lambda D}{\Delta y} = \frac{2 \times 6000 \times 10^{-10} \times 0.50}{3 \times 10^{-3}} = 0.2\text{ mm}\).
If \( \beta_1, \beta_2 \) and \( \beta_3 \) are fringe widths for red, green and violet light respectively for the same setup of Young’s double slit experiment, then
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 \( \beta_1 > \beta_2 > \beta_3 \).
A screen is placed \(50\text{ cm}\) from a single slit, which is illuminated with \(6000\text{ A^0}\) light. If distance between the first and third minima in the diffraction pattern is \(3\text{ mm}\), the width of the slit is
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 \(d = 2 \times 10^{-4}\text{ m} = 0.2\text{ mm}\).
Assertion (A): Resolving power of a microscope is different for different colours of illuminating light.
Reason (R): Resolving power of a microscope is directly proportional to the wavelength of illuminating light.
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.
Assertion (A): When a width of one of the slits of Young’s double slit experiment is double that of the other than brighter fringes are nine times brighter than the dark fringes.
Reason (R): The amplitude of the wave is proportional to the width of the slit.
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): When tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the centre of shadow of the obstacle.
Reason (R): Constructive interference occurs at the centre of the shadow.
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.
Assertion (A): If width of one of the slit in YDSE is slightly increased, then maximum and minimum both Intensity will increase.
Reason (R): Intensity reaching from that slit on screen will slightly increase.
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.
Assertion (A): If white light is used in place of monochromatic light in YDSE then central point is white. Although at other places coloured fringes will be obtained.
Reason (R): At centre path difference is zero for all wave lengths. Hence all wave will interfere constructively.
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.