In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(= 1.5), the central fringe shifts by 0.2 cm. The value of t is ____ cm.

Related Formula

Core Logic

The angular separation (angular fringe width) θ$\theta$ is given by (λ)/(d)$\frac{\lambda}{d}$, where λ$\lambda$ is the wavelength and d$d$ is the separation between slits. It is completely independent of the distance to the screen D$D$.

Statement I claims moving the screen away (increasing D$D$) increases angular separation. Since θ$\theta$ doesn't depend on D$D$, Statement I is false.

Statement II claims using a higher wavelength λ$\lambda$ increases angular separation. Since θ ∝ λ$\theta \propto \lambda$, increasing λ$\lambda$ increases θ$\theta$. Statement II is true.

Step 1: Final Conclusion

Statement I is false but Statement II is true.

Pattern Recognition

Linear fringe width y = λ D / d$y = \lambda D / d$ depends on D$D$, but angular fringe width θ = λ / d$\theta = \lambda / d$ isolates only the slit separation and wavelength.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

Core Logic

Introducing a transparent sheet of thickness t$t$ and refractive index μ$\mu$ adds an extra optical path length of (μ - 1)t$(\mu - 1)t$. This translates to a shift of the entire fringe pattern on the screen. Given values: d = 0.4 mm = 0.4 × 10⁻³ m$d = 0.4 \text{ mm} = 0.4 \times 10^{-3} \text{ m}$ D = 1 m$D = 1 \text{ m}$ t = 20 = 20 × 10⁻⁶ m$t = 20 \,\mu\text{m} = 20 \times 10^{-6} \text{ m}$ yshift = 20 mm = 20 × 10⁻³ m$y_{\text{shift}} = 20 \text{ mm} = 20 \times 10^{-3} \text{ m}$

Step 1: Substitution and Solving

Step 2: Final Conclusion

We are given μ = (α)/(10)$\mu = \frac{\alpha}{10}$.

Pattern Recognition

The central fringe always shifts towards the slit covered by the slab. The linear shift y₀ = (D)/(d)(μ-1)t$y_0 = \frac{D}{d}(\mu-1)t$ is independent of the wavelength used.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

Core Logic

Since frequency f$f$ remains constant across different media:

Given μ₁ = (4)/(3), λ₁ = 540 ~nm$\mu_1 = \frac{4}{3}, \lambda_1 = 540 \mathrm{~nm}$ and μ₂ = (3)/(2)$\mu_2 = \frac{3}{2}$:

Step 1: Final Conclusion

The wavelength in the second medium is 480 ~nm$480 \mathrm{~nm}$.

Pattern Recognition

Product rule: μ · λ = constant$\mu \cdot \lambda = \text{constant}$. (4/3) × 540 = (3/2) × λ₂ λ₂ = 480 ~nm$(4/3) \times 540 = (3/2) \times \lambda_2 \implies \lambda_2 = 480 \mathrm{~nm}$.

Chapter Mix

Class 12 Physics: Wave Optics

Related Formula

Core Logic

Analyzing each statement using width of central maxima formula βcm = (2λ D)/(a)$\beta_{\text{cm}} = \frac{2\lambda D}{a}$:

• (A) βcm ∝ λ$\beta_{\text{cm}} \propto \lambda$: Width increases with increasing λ$\lambda$. (Correct)
• (B) Incorrect since βcm$\beta_{\text{cm}}$ decreases when λ$\lambda$ decreases.
• (C) βcm ∝ (1)/(a)$\beta_{\text{cm}} \propto \frac{1}{a}$: Width increases with decreasing slit width a$a$. (Correct)
• (D) Incorrect.
• (E) Intensity/brightness of central maxima increases as λ$\lambda$ decreases due to reduced diffraction spread. (Correct)

Statements A, C, and E are correct.

Step 1: Final Conclusion

The correct combination of statements is A, C & E. (NTA Official Answer Key specifies Option 1).

Pattern Recognition

Single slit width: βcm ∝ λ/a$\beta_{cm} \propto \lambda/a$. Central maxima widens with larger wavelength or smaller slit width.

Chapter Mix

Class 12 Physics: Wave Optics