Ray Optics JEE Main Questions Available - Important PYQs with Solutions

Importance Of Ray Optics JEE Mains Questions

1. It has high weightage as previous year questions with solutions on optics are expected to be 2-3 every year in JEE Mains paper.

2. It is scoring as questions are mostly direct and application based.

3. It forms the foundation of advanced topics like reflection and refraction so which makes it more important.

4. There is a real life application, and it is also required in practical exams.

Ray Optics JEE Main Previous Year Questions With Solutions PDF

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Previous Year Questions with Solutions on Ray Optics Important Topics

This table has the topics of Ray Optics JEE Mains and what their key focus areas are:

Ray Optics PYQ JEE mains 2026

We have listed below few Ray Optics PYQ JEE Mains for your reference:

Question 1:If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30∘ in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is μm..

Solution:

Fraunhofer diffraction by a single slit

Let's assume a plane wave front is incident on a slit AB (of width b). Each and every part of the exposed part of the plane wave front (i.e. every part of the slit) acts as a source of secondary wavelets spreading in all directions. The diffraction is obtained on a screen placed at a large distance. (In practice, this condition is achieved by placing the screen at the focal plane of a converging lens placed just after the slit).

• The diffraction pattern consists of a central bright fringe (central maxima) surrounded by dark and bright lines (called secondary minima and maxima).

• At point O on the screen, the central maxima is obtained. The wavelets originating from points A and B meets in the same phase at this point, hence at O, intensity is maximum

Secondary minima : For obtaining nth secondary minima at P on the screen, path difference between the diffracted waves

Δx=bsin⁡θ=nλ

1. Angular position of nth secondary minima:

sin⁡θ≈θ=nλb

2. Distance of nth secondary minima from central maxima:

xn=D⋅θ=nλDb≈D=D= Focal length of converging lens.

Secondary maxima: For nth secondary maxima at P on the screen.
Path difference

Δx=bsin⁡θ=(2n+1)λ2; wheren =1,2,3……

(i) Angular position of nth secondary maxima

sin⁡θ≈θ≈(2n+1)λ2b

(ii) Distance of nth secondary maxima from central maxima:

xn=D⋅θ=(2n+1)λD2b

θ1=sin−1⁡(2λa)θ2=sin−1⁡(3λa)∵θ1+θ2=30∘

⇒sin−1⁡(2λa)+sin−1⁡(3λa)=π6⇒2λa1−(3λa)2+3λa1+(2λa)2=sin⁡π6

Here λ=628 nm
After solving

A=6.07μ m

Approximate Method :

θ=θ1+θ2⇒π6=2λa+3λa⇒π6=5a(628 nm)⇒a=6μ m

Hence, the answer is 6.

Question 2: The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm . The refractive index of the lens material is:

1) 1.2

2) 1.4

3) 1.5

4) 1.8

Solution:

The lens maker's formula relates the focal length f of a lens to the refractive index μ and the radii of curvature of the two surfaces of the lens. The formula is given by:

1f=(μ−1)(1R1−1R2)

Where:
- f is the focal length of the lens,
- μ is the refractive index of the lens material,
- R1 and R2 are the radii of curvature of the two surfaces of the lens.

1f=(μ−1)(1R1−1R2)112=(μ−1)(110−1−15)112=(μ−1)(3+230)μ=32

Hence, the answer is the option (3)

Question 3: In a Young's double slit experiment, the slits are separated by 0.2 mm. If the slits separation is increased to 0.4 mm, the percentage change of the fringe width is:
1) 0%

2)100%
3) 50%

4) 25%

Solution:

Band Width (β) -

The distance between any two consecutive bright or dark bands is called bandwidth.

Take the consecutive dark or bright fringe -

xn+1−xn=(n+1)λDd−(n)λDdxn+1−xn=λDdβ=λDd
Angular fringe width -

θ=βD=λD/dD=λd

β=Dλ d∝1 d

If d is doubled then β is half so 50% decrement.

Hence, the answer is the option (3).

Question 4: The width of one of the two slits in Youngs double-slit experiment is d, while that of the other slit is xd. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9:4, then what is the value of x?

(Assume that the field strength varies according to the slit width.)

1) 2

2) 3

3) 4

4) 5

Solution:

Young's double slit experiment- 2 -

Intensity of Fringes In Young’s Double Slit Experiment-

For two coherent sources S₁ and S₂, the resultant intensity at point P on the screen is given by.

I=I1+I2+2I1I2cos⁡ϕ

where
I1= The intensity of wave from S1
I2= The intensity of wave S2
Putting I1 and I2=Io (Because d≪<D )

⇒I=I0+I0+2I0I0cos⁡ϕ=4I0cos2⁡ϕ2

So the intensity variation from maximum to minimum depends on the phase difference. Let us discuss one by one -

For maximum intensity
The phase difference between the waves at the point of observation is ϕ=0∘ or 2nπ. Path difference between the waves at the point of observation is Δx=nλ( i.e. even multiple of λ/2 )

The resultant intensity at the point of observation will be the maximum.

i.e Imax=I1+I2+2I1I2Imax=(I1+I2)2 If I1=I2=I0⇒Imax=4I0

For Minimum Intensity -
The phase difference between the waves at the point of observation is

ϕ=180∘ or (2n−1)π;n=1,2,… or (2n+1)π;n=0,1,2…

Path difference between the waves at the point of observation is

Δx=(2n−1)λ2( i.e. odd multiple of λ/2)

The resultant intensity at the point of observation will be the maximum.

Imin=I1+I2−2I1I2Imin=(I1−I2)2 If I1=I2=I0⇒Imin=0

I∝( width )2(I1+I2I1−I2)2=94I1+I2I1−I2=32(x+1)d(x−1)d=32⇒3x−3=2x+2x=5

Hence, the answer is the option (3).

Question 5: Youngs double-slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is:-

1) 4

2) 8

3) 6

4) 5

Youngs double-slit experiment -

This experiment is performed by British physicist Thomas Young. He used an arrangement as shown below. In this he used a monochromatic source of light S. He made two pinholes S₁ and S₂ (very close to each other) on an opaque screen as shown in the figure Each source can be considered as a source of the coherent light source.

Bright Fringes:

By the principle of interference, the condition for constructive interference is the path difference = n&lambda

xdD=nλ

Here, n=0,1,2…… indicate the order of bright fringes

So,x=(nλDd)

This equation gives the distance of the nth bright fringe from the point O.

For Dark fringes :

By the principle of interference, the condition for destructive interference is the path difference =

(2n−1)λ2

Here, n=1,2,3… indicate the order of the dark fringes.

So,x=(2n−1)λD2d

The above equation gives the distance of the nth dark fringe from the point O .

So, we can say that the alternate dark and bright fringe will be obtained on either side of the central bright fringe.

The condition for coincidence of fringes is given as,

nλ1Dd=mλ2Ddn480=m6004n=5mnmin=5

Hence, the answer is the option (4).