Home
Class 12
PHYSICS
Two slits S(1) and S(2) illuminated by a...

Two slits `S_(1)` and `S_(2)` illuminated by a white light source give a white central maxima. A transparent sheet of refractive index 1.25 and thickness `t_(1)` is placed in front of `S_(1)`. Another transparent sheet of refractive index 1.50 and thickness `t_(2)` is placed in front of `S_(2)`. If central maxima is not effected, then ratio of the thickness of the two sheets will be :

A

`1:2`

B

`2:1`

C

`1:4`

D

`4:1`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the ratio of the thickness of two transparent sheets placed in front of two slits, such that the central maxima remains unaffected. ### Step-by-Step Solution: 1. **Understanding Optical Path Length**: The optical path length (OPL) is given by the product of the refractive index (μ) of the medium and the actual path length (t). The formula for optical path length is: \[ \text{OPL} = \mu \times t \] 2. **Identifying the Given Values**: - For the first slit \( S_1 \): - Refractive index \( \mu_1 = 1.25 \) - Thickness \( t_1 \) - For the second slit \( S_2 \): - Refractive index \( \mu_2 = 1.50 \) - Thickness \( t_2 \) 3. **Calculating the Optical Path Lengths**: - The optical path length for the first sheet in front of \( S_1 \): \[ \text{OPL}_1 = \mu_1 \times t_1 = 1.25 t_1 \] - The optical path length for the second sheet in front of \( S_2 \): \[ \text{OPL}_2 = \mu_2 \times t_2 = 1.50 t_2 \] 4. **Condition for Central Maxima**: For the central maxima to remain unaffected, the optical path lengths must be equal: \[ \text{OPL}_1 = \text{OPL}_2 \] Therefore, we have: \[ 1.25 t_1 = 1.50 t_2 \] 5. **Rearranging the Equation**: To find the ratio of the thicknesses \( \frac{t_1}{t_2} \): \[ \frac{t_1}{t_2} = \frac{1.50}{1.25} \] 6. **Simplifying the Ratio**: \[ \frac{t_1}{t_2} = \frac{1.50 \div 1.25}{1.25 \div 1.25} = \frac{1.50 \div 1.25}{1} = \frac{1.50}{1.25} = \frac{6}{5} = \frac{12}{10} = \frac{2}{1} \] 7. **Final Result**: The ratio of the thickness of the two sheets is: \[ \frac{t_1}{t_2} = 2:1 \] ### Conclusion: Thus, the ratio of the thickness of the two sheets \( t_1 \) and \( t_2 \) is \( 2:1 \).
Promotional Banner

Similar Questions

Explore conceptually related problems

Find Brewster.s angle for a transparent liquid having refractive index 1.5.

In YDSE when slab of thickness t and refractive index mu is placed in front of one slit then central maxima shifts by one fringe width. Find out t in terms of lambda and mu .

To make the central fringe at the center O , mica sheet of refractive index 1.5 is introduced Choose the corect statement.

Young's double slit experiment is conducted in a liquid of refractive index mu_1 as shown in figure. A thin transparent slab of refractive index mu_2 is placed in front of the slit s_2 . The magnitude of optical path difference at 'O' is

Phase difference at the central point changes by pi//3 when as thick film having a refractive index 1.5 and thickness 0.4mum is placed in front of upper slit of a YDSE set up. If the wavelength (in nm) of the light used is 600 k, find k.

In YDSE arrangement as shown in figure, fringes are seen on screen using monochromatic source S having wavelength 3000 Å (in air). S_1 and S_2 are two slits seperated by d = 1 mm and D = 1m. Left of slits S_1 and S_2 medium of refractive index n_1 = 2 is present and to the right of S_1 and S_2 medium of n_2 = 3/2 , is present. A thin slab of thickness 't' is placed in front of S_1 . The refractive index of n_3 of the slab varies with distance from it's starting face as shown in figure. In order to get central maxima at the centre of screen, the thickness of slab required is :

In figure S is a monochromatic source of light emitting light of wavelength of wavelength lambda (in air). Light on slits S_(1) from S and then reaches in the slit S_(2) and S_(3) through a medium of refractive index mu_(1) . Light from slit S_(2) and S_(3) reaches the screen through a medium of refractive index mu_(3) . A thin transparent film of refractive index mu_(2) and thickness t is used placed in front of S_(2) . Point P is symmetrical w.r.t. S_(2) and S_(3) . Using the values d = 1 mm, D = 1 m, mu_(1) = 4//3 , mu_(2) = 3//2, mu_(3) = 9//5 , and t = (4)/(9) xx 10^(-5) m , a. find distance of central maxima from P, b. If the film in front of S_(2) is removed, then by what distance and in which direction will be central maxima shift ? .

In Young's double slit experiment the slits, S_(1) & S_(2) are illuminated by a parallel beam of light of wavelength 4000 Å from the medium of refractive index n_(1)=1.2 A thin film of thickness 1.2 mum and refractive index n=1.5 is placed infrom of S_(1) perpedicular to path of light. the refractive index of medium between plane of slits & screen is n_(2)=1.4 if the light coming from the film and S_(1)&S_(2) have equal intensities I then intensity at geometrical centre of the screen O is

In Young's double-slit experiment, a point source is placed on a solid slab of refractive index 6//5 at a distance of 2 mm from two slits spaced 3 mm apart as shown and at equal distacne from both the slits. The screen is at a distance of 1 m from the slits. Wavelength of light used is 500 nm. a. Find the position of the central maximum. b. Find the order of the fringe formed at O. c. A film of refractive index 1.8 is to be placed in front of S_(1) so that central maxima is formed where 200th maxima was formed. Find the thickness of film.

In YDSE , slab of thickness t and refractive index mu is placed in front of any slit. Then displacement of central maximu is terms of fringe width when light of wavelength lamda is incident on system is