Wave Optics Set-1

Q1. In Young's double-slit experiment the slits are separated by $d=0.50\ \text{mm}$ and the screen is at a distance $D=2.0\ \text{m}$. Monochromatic light of wavelength $\lambda=600\ \text{nm}$ is used. Using the small-angle approximation the fringe width $\beta$ is $\beta=\dfrac{\lambda D}{d}$. Numerically, $\beta$ equals:

Q2. In a Young's double-slit arrangement $d=0.20\ \text{mm}$, $D=1.20\ \text{m}$, and $\lambda=600\ \text{nm}$. A thin glass slab of refractive index $n=1.50$ is placed in front of one slit and the central bright fringe shifts by $2.4\ \text{mm}$ toward that slit. Using $m=\dfrac{x}{\beta}$ and $m=\dfrac{(n-1)t}{\lambda}$, the thickness $t$ of the slab is closest to:

Q3. Assertion: For a thin film of refractive index $n>1$ in air, the condition for constructive interference in reflected light is $2nt=\bigl(m+\tfrac{1}{2}\bigr)\lambda$ (integer $m$).
Reason: One of the two reflected rays suffers a phase change of $\pi$ upon reflection while the other does not; hence an extra half‑wavelength effective phase difference appears.

Q4. Monochromatic light $\lambda=600\ \text{nm}$ illuminates two identical narrow slits each of width $a$ and with center separation $d$. On a screen at $D=2.0\ \text{m}$ the distance between the first minima of the single‑slit diffraction envelope on either side of the central maximum is $6.0\ \text{mm}$, and the adjacent interference bright fringes (inside the envelope) are spaced by $2.0\ \text{mm}$. Using $a\sin\theta=\lambda$ (minima) and $\beta=\dfrac{\lambda D}{d}$ (fringe spacing), the values of $a$ and $d$ are approximately:

Q5. A source with central wavelength $\lambda=600\ \text{nm}$ and spectral width $\Delta\lambda=3\ \text{nm}$ illuminates two slits separated by $d=0.40\ \text{mm}$. The screen is at $D=1.50\ \text{m}$. Using the approximate coherence length $L_c\approx\dfrac{\lambda^2}{\Delta\lambda}$ and that the path difference at a screen point $x$ is $\approx \dfrac{d\,x}{D}$, estimate the lateral extent (on one side from centre) over which fringes remain visible (i.e., solve $d\,x/D\approx L_c$):

Q6. In a Michelson interferometer monochromatic light of wavelength $\lambda=500\ \text{nm}$ is used. A thin transparent plate of refractive index $n=1.6$ is inserted normally in one arm and its effective optical thickness is increased (beam passes twice through the plate) causing 100 fringe shifts. If $N$ is the number of fringes shifted and $N=\dfrac{2(n-1)t}{\lambda}$, the increase in physical thickness $t$ of the plate is approximately:

Q7. Assertion: If the amplitude of light from one slit is halved while the other remains unchanged, the fringe visibility (contrast) in the interference pattern decreases.
Reason: Visibility $V$ depends only on the phase difference between the two waves and is independent of their amplitudes.

Q8. Two very close wavelengths $\lambda_1=600\ \text{nm}$ and $\lambda_2=603\ \text{nm}$ illuminate a double‑slit of separation $d=0.50\ \text{mm}$. The screen is at $D=2.0\ \text{m}$. Successive positions where bright fringes from both wavelengths coincide are separated approximately by

The distance $L$ is about:

Q9. Assertion: In Newton's rings observed in reflected light the central spot is dark.
Reason: At the point of contact the air film thickness is effectively zero and one of the reflected rays suffers a phase inversion of $\pi$ while the other does not, producing destructive interference at the centre.

Q10. A single slit of width $a=0.20\ \text{mm}$ is illuminated by light of wavelength $\lambda=600\ \text{nm}$. A screen is placed at $D=1.0\ \text{m}$. The linear width (distance between first minima on either side) of the central diffraction maximum is given approximately by $w=2\lambda D/a$. Numerically $w$ is: