Ray Optics JEE Main Questions Available - Important PYQs with Solutions

Why Solve Ray Optics Questions or JEE Mains 2027?

1. It has high weightage, as previous years questions with solutions on optics are expected to be 2-3 every year in the JEE Main exam.

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

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

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

Download Ray Optics JEE Main Previous Year Questions PDF

Ray Optics is one of the most consistently asked topics in JEE Main, making it a must-prepare chapter for every aspirant. Go through these previous year questions and solutions to understand the pattern and sharpen your problem-solving approach.

Important Topics Covered in Ray Optics for JEE Main

Ray Optics is a broad chapter that includes several important concepts frequently tested in JEE Main. Topics such as reflection, refraction, lenses, mirrors, optical instruments, interference, and polarization form the foundation of many numerical and conceptual questions. The table provided below highlights the important topics and their key focus areas for JEE Main preparation:

Ray Optics JEE Main Previous Year Questions 2027

One of the best ways to prepare Ray Optics for JEE Main 2027 is practising previous year questions. These questions help students to get familiarised with the exam pattern and repeated concepts and to increase the speed and accuracy in solving problems. Here are some important PYQs on Ray Optics with detailed solutions:

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 $n$th secondary minima at P on the screen, path difference between the diffracted waves:

$\Delta x = b\sin\theta = n\lambda$

1. Angular position of $n$th secondary minima:

$\sin\theta \approx \theta = \dfrac{n\lambda}{b}$

2. Distance of $n$th secondary minima from central maxima:

$x_n = D \cdot \theta = \dfrac{n\lambda D}{b}$

where $D$ = Focal length of converging lens.

Secondary maxima: For $n$th secondary maxima at P on the screen, the path difference:

$\Delta x = b\sin\theta = \dfrac{(2n+1)\lambda}{2}$ ; where $n = 1, 2, 3\ldots$

(i) Angular position of $n$th secondary maxima:

$\sin\theta \approx \theta \approx \dfrac{(2n+1)\lambda}{2b}$

(ii) Distance of $n$th secondary maxima from central maxima:

$x_n = D \cdot \theta = \dfrac{(2n+1)\lambda D}{2b}$

$\theta_1 = \sin^{-1}\left(\dfrac{2\lambda}{a}\right), \quad \theta_2 = \sin^{-1}\left(\dfrac{3\lambda}{a}\right)$

$\because \theta_1 + \theta_2 = 30^\circ$

$\Rightarrow \sin^{-1}\left(\dfrac{2\lambda}{a}\right) + \sin^{-1}\left(\dfrac{3\lambda}{a}\right) = \dfrac{\pi}{6}$

$\Rightarrow \dfrac{2\lambda}{a}\sqrt{1-\left(\dfrac{3\lambda}{a}\right)^2} + \dfrac{3\lambda}{a}\sqrt{1-\left(\dfrac{2\lambda}{a}\right)^2} = \sin\dfrac{\pi}{6}$

Here $\lambda = 628$ nm, after solving:

$a = 6.07\ \mu m$

Approximate Method:

$\theta = \theta_1 + \theta_2 \Rightarrow \dfrac{\pi}{6} = \dfrac{2\lambda}{a} + \dfrac{3\lambda}{a}$

$\Rightarrow \dfrac{\pi}{6} = \dfrac{5}{a}(628\ \text{nm})$

$\Rightarrow a = 6\ \mu m$

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 $\mu$ and the radii of curvature of the two surfaces of the lens. The formula is given by:

$\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

Where:

• $f$ is the focal length of the lens
• $\mu$ is the refractive index of the lens material
• $R_1$ and $R_2$ are the radii of curvature of the two surfaces of the lens

$\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$

$\dfrac{1}{12} = (\mu - 1)\left(\dfrac{1}{10} - \dfrac{1}{-15}\right)$

$\dfrac{1}{12} = (\mu - 1)\left(\dfrac{3 + 2}{30}\right)$

$\dfrac{1}{12} = (\mu - 1)\left(\dfrac{5}{30}\right)$

$\dfrac{1}{12} = (\mu - 1)\dfrac{1}{6}$

$\mu - 1 = \dfrac{1}{2}$

$\mu = \dfrac{3}{2}$

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 -

$x_{n+1} - x_n = \dfrac{(n+1)\lambda D}{d} - \dfrac{(n)\lambda D}{d}$

$x_{n+1} - x_n = \dfrac{\lambda D}{d}$

$\beta = \dfrac{\lambda D}{d}$

Angular fringe width:

$\theta = \dfrac{\beta}{D} = \dfrac{\lambda D/d}{D} = \dfrac{\lambda}{d}$

$\beta = \dfrac{D\lambda}{d}$

$\beta \propto \dfrac{1}{d}$

If $d$ is doubled then $\beta$ 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-

Here is the corrected LaTeX in single dollar signs:

For two coherent sources $S_1$ and $S_2$, the resultant intensity at point P on the screen is given by:

$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$

where $I_1$ = The intensity of wave from $S_1$, $I_2$ = The intensity of wave from $S_2$.

Putting $I_1$ and $I_2 = I_0$ (Because $d \ll D$):

$\Rightarrow I = I_0 + I_0 + 2\sqrt{I_0 I_0}\cos\phi = 4I_0\cos^2\dfrac{\phi}{2}$

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 $\phi = 0^\circ$ or $2n\pi$.

Path difference between the waves at the point of observation is $\Delta x = n\lambda$ (i.e. even multiple of $\lambda/2$).

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

$I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2}$

$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$

If $I_1 = I_2 = I_0 \Rightarrow I_{max} = 4I_0$

For Minimum Intensity:

The phase difference between the waves at the point of observation is:

$\phi = 180^\circ$ or $(2n-1)\pi;\ n = 1, 2, \ldots$ or $(2n+1)\pi;\ n = 0, 1, 2\ldots$

Path difference between the waves at the point of observation is:

$\Delta x = \dfrac{(2n-1)\lambda}{2}$ (i.e. odd multiple of $\lambda/2$)

The resultant intensity at the point of observation will be minimum:

$I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2}$

$I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$

If $I_1 = I_2 = I_0 \Rightarrow I_{min} = 0$

$I \propto (\text{width})^2$

$\left(\dfrac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^2 = \dfrac{9}{4}$

$\dfrac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} = \dfrac{3}{2}$

$\dfrac{(\sqrt{x}+1)d}{(\sqrt{x}-1)d} = \dfrac{3}{2}$

$\Rightarrow 3\sqrt{x} - 3 = 2\sqrt{x} + 2$

$\sqrt{x} = 5$

$x = 5$

Hence, the answer is 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 $= n\lambda$

$\dfrac{xd}{D} = n\lambda$

Here, $n = 0, 1, 2\ldots$ indicate the order of bright fringes.

So, $x = \dfrac{n\lambda D}{d}$

This equation gives the distance of the $n$th bright fringe from the point O.

For Dark Fringes:

By the principle of interference, the condition for destructive interference is the path difference $= \dfrac{(2n-1)\lambda}{2}$

Here, $n = 1, 2, 3\ldots$ indicate the order of the dark fringes.

So, $x = \dfrac{(2n-1)\lambda D}{2d}$

The above equation gives the distance of the $n$th dark fringe from the point O.

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

The condition for coincidence of fringes is given as:

$\dfrac{n\lambda_1 D}{d} = \dfrac{m\lambda_2 D}{d}$

$\dfrac{n}{480} = \dfrac{m}{600}$

$4n = 5m$

$n_{min} = 5$

Hence, the answer is Option (4).