In a Young's double slit experiment, three polarizers are kept as shown in the figure . The transmission axes of P1 and…

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Q36 jee_main_2026_21_jan_morning Interference

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.

A. $8 \times 10⁻⁴$ $8 \times 10^{-4}$
B. $6.0 \times 10⁻³$ $6.0 \times 10^{-3}$
C. $5.6 \times 10⁻⁴$ $5.6 \times 10^{-4}$
D. $5.0 \times 10⁻³$ $5.0 \times 10^{-3}$

Solution

Related Formula

Shift (x) = (D)/(d) (μ - 1)t

Alternatively written: $d ((x)/(D)) = (μ - 1)t$

Core Logic

Given parameters: Distance between slits, $d = 0.1 cm$ $d = 0.1\text{ cm}$ Distance to screen, $D = 50 cm$ $D = 50\text{ cm}$ Fringe shift, $x = 0.2 cm$ $x = 0.2\text{ cm}$ Refractive index, $μ = 1.5$ $\mu = 1.5$

Step 1: Calculate Thickness

Rearranging the formula for thickness $t$ $t$:

t = (x \cdot d)/(D(μ - 1))

t = ((0.2)(0.1))/(50(1.5 - 1))

t = (0.02)/(50(0.5)) = (0.02)/(25)

t = 8 \times 10⁻⁴ cm

Pattern Recognition

Direct plug-and-play into the slab shift formula. Working completely in cm avoids unit conversion errors as long as all given lengths ( $x$ $x$, $d$ $d$, $D$ $D$) are consistently in cm and the answer is requested in cm.

Chapter Mix

Class 12 Physics: Wave Optics

Q34 jee_main_2026_21_jan_evening Young's Double Slit Experiment

Given below are two statements: Statement I : In a Young's double slit experiment, the angular separation of fringes will increase as the screen is moved away from the plane of the slits Statement II : In a Young's double slit experiment, the angular separation of fringes will increase when monochromatic source is replaced by another monochromatic source of higher wavelength In the light of the above statements, choose the correct answer from the options given below:

A. $Both Statement I and Statement II are true$
B. $Both Statement I and Statement II are false$
C. $Statement I is false but Statement II is true$
D. $Statement I is true but Statement II is false$

Solution

Related Formula

Angular fringe width θ = (λ)/(d)

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 $θ \propto λ$ $\theta \propto \lambda$, 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

Q49 jee_main_2026_21_jan_evening YDSE - Fringe Shift

In a Young's double slit experiment set up, the two slits are kept

0.4 mm

apart and screen is placed at

1 m

from slits. If a thin transparent sheet of thickness

20

is introduced in front of one of the slits then centre bring fringe shifts by

20 mm

on the screen. The refractive index of transparent sheet is given by

(α)/(10)

, where

α

is ________.

Numerical Answer. Answer: 14 to 14

Solution

Related Formula

yshift = ((μ - 1)t \cdot D)/(d)

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: $d = 0.4 mm = 0.4 \times 10⁻³ m$ $D = 1 m$ $D = 1 \text{ m}$ $t = 20 = 20 \times 10⁻⁶ m$ $yshift = 20 mm = 20 \times 10⁻³ m$

Step 1: Substitution and Solving

20 \times 10⁻³ = (μ - 1) \times 20 \times 10⁻⁶ \times 10.4 \times 10⁻³

20 \times 10⁻³ = (μ - 1) \times 20 \times 10⁻⁶4 \times 10⁻⁴

20 \times 10⁻³ = (μ - 1) \times 5 \times 10⁻²

(μ - 1) = 20 \times 10⁻³5 \times 10⁻² = 4 \times 10⁻¹ = 0.4

Step 2: Final Conclusion

μ = 1 + 0.4 = 1.4

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

(α)/(10) = 1.4 α = 14

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

Q37 jee_main_2026_22_january_evening Refractive Index and Wavelength

The wavelength of light, while it is passing through water is 540 nm. The refractive index of water is

(4)/(3)

. The wavelength of the same light when it is passing through a transparent medium having refractive index of

(3)/(2)

is ____ nm.

A. $380$ $380$
B. $840$ $840$
C. $480$ $480$
D. $540$ $540$

Solution

Related Formula

μ = (c)/(v) = (λ₀)/(λ)

μ₁ λ₁ = μ₂ λ₂

Core Logic

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

(μ₁)/(μ₂) = (λ₂)/(λ₁)

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

(4/3)/(3/2) = (λ₂)/(540)

λ₂ = ((4 \times 2)/(3 \times 3)) \times 540 = (8)/(9) \times 540 = 480 ~nm

Step 1: Final Conclusion

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

Pattern Recognition

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

Chapter Mix

Class 12 Physics: Wave Optics

Q40 jee_main_2026_22_january_evening Single Slit Diffraction

Which of the following are true for a single slit diffraction? (A) Width of central maxima increases with increase in wavelength keeping slit width constant. (B) Width of central maxima increases with decrease in wavelength keeping slit width constant. (C) Width of central maxima increases with decrease in slit width at constant wavelength. (D) Width of central maxima increases with increase in slit width at constant wavelength. (E) Brightness of central maxima increases for decrease in wavelength at constant slit width. Options :

A. A, D, E only
B. A, D only
C. B, D only
D. B, C only

Solution

Related Formula

βcm = (2λ D)/(a)

Core Logic

Analyzing each statement using width of central maxima formula $βcm = (2λ D)/(a)$ :

(A) $βcm \propto λ$ $\beta_{\text{cm}} \propto \lambda$: Width increases with increasing $λ$ $\lambda$. (Correct)
(B) Incorrect since $βcm$ $\beta_{\text{cm}}$ decreases when $λ$ $\lambda$ decreases.
(C) $βcm \propto (1)/(a)$ $\beta_{\text{cm}} \propto \frac{1}{a}$: 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: $βcm \propto λ/a$ $\beta_{cm} \propto \lambda/a$. Central maxima widens with larger wavelength or smaller slit width.

Chapter Mix

Class 12 Physics: Wave Optics

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