Previous Year Questions with Solutions on Optics

1.0Introduction

Ray Optics treats light as rays traveling in straight lines, explaining basic phenomena like reflection and refraction using principles such as Snell's Law. It’s useful for understanding lenses, mirrors, and prisms. Wave Optics, on the other hand, views light as an electromagnetic wave, focusing on its wave properties like interference, diffraction, and polarization. It helps explain more complex behaviors of light, like those in microscopes and diffraction gratings. Ray Optics simplifies basic optical systems, while Wave Optics addresses more advanced wave-related phenomena.

Optics is crucial for JEE Main, covering ray optics (lenses, mirrors) and wave optics (interference, diffraction, polarization). It builds a strong foundation for advanced physics and real-world applications like telescopes. Focus on understanding concepts, practicing problems, using diagrams, studying NCERT, and revising past papers to excel.

Note:For JEE Main, expect 2-3 questions on optics, covering reflection, refraction, lenses, and optical instruments, as well as wave optics topics like Huygens' principle, interference, diffraction, and polarization. This will contribute to around 3% of the total marks (12 marks). Focus on path difference, phase difference, and interference.

2.0Key Concepts to Remember

Definitions

1.Condition for rectilinear propagation of Light

1.The medium is isotropic.

2.The obstacle or opening through which light passes is not very small.

3.Bending is negligible if $\frac{D\lambda }{a}\ll a\text{ or }a\gg \sqrt{D\lambda }$. If this condition is fulfilled, light is

said to move rectilinearly.

2.Plane Mirror

• A plane mirror is formed by polishing one surface of a plane thin glass plate. It is also said to be silver on one side.
• A beam of parallel light rays incident on a plane mirror will reflect as a parallel beam of rays.

3.Spherical Mirrors

• A curved mirror is part of a hollow sphere. If reflection takes place from the inner surface then the mirror is called concave and if its outer surface acts as reflector it is convex.

4.Refraction of Light

Refraction occurs when light changes mediums, causing a shift in speed and direction. If light strikes at an angle (0° < i < 90°) {(\0\circ < i < \90\circ), it bends due to this speed change. Light incident normally passes straight but is still refracted. Refraction without reflection is impossible, and as the angle of incidence increases, more energy is reflected. The refractive index is the ratio of light speed in vacuum to its speed in the medium.

5.Critical Angle and Total Internal Reflection ( T. I. R.)

The critical angle is the angle in the denser medium where the refraction in the rarer medium is 90° {\ 90\circ}. If the angle exceeds the critical angle, total internal reflection occurs, and the interface acts like a mirror.

6.Prism

"A prism is a homogeneous, solid, transparent refracting medium with two plane surfaces inclined at an angle."

7. Lens Theory

A lens is a transparent object with two refracting surfaces, at least one of which is curved, and the refractive index of the material differs from that of its surrounding medium.

8.Wavefront and Types of Wavefront

The Locus of all particles vibrating in the same phase is called the wavefront; its shape depends upon the shape of the light source from which it originates.

9. Young's Double Slit Experiment.(YDSE)

Light wave nature is proved experimentally by YDSE,in this experiment division of wavefront takes place.

10.Diffraction

The bending of light rays from the sharp edges of an opaque obstacle or aperture and its spreading in the geometrical shadow region is called diffraction.

11. Polarisation of Light Wave

• It is the phenomenon of constraining the vibration of light (electric vector) to a specific direction perpendicular to the wave's direction of propagation.
• Polarisation confirms the transverse nature of waves.

Deviation by two plane mirrors	$\delta =360^{\circ }-2\theta$
Relation between velocity of object and image	For x-Axis $v_{iG}-v_{mG}=-\left(v_{oG}-v_{mG}\right)$ for y axis and z axis $v_{iG}-v_{mG}=\left(v_{oG}-v_{mG}\right)\text{ or }{v}_{iG}=v_{oG}$ $v_{iG}=\text{velocity of image with respect to ground}.$
Number of images formed by two inclined mirrors	(1) $\frac{360^{\circ }}{\theta }=\text{even number;}\text{number of image}=\frac{360^{\circ }}{\theta }-1$ (2)If $\frac{360^{\circ }}{\theta }=\text{odd number;}\text{number of image}=\frac{360^{\circ }}{\theta }-1$ if the object is placed on the angle bisector. (3) If $\frac{360^{\circ }}{\theta }=\text{odd number;}\text{number of image}=\frac{360^{\circ }}{\theta }$, if the object is not placed on the angle bisector. (4) If $360°\ne Integer\frac{360^{\circ }}{\theta }=\text{Integer}$, then count the number of images as explained above.
Mirror formula	$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Lateral Magnification	$m=\frac{h_i}{h_o}=-\frac{v}{u}$
Longitudinal magnification	$m_L=\frac{\text{Length of Image}}{\text{Length of Object}}=\left|\frac{v_2-v_1}{u_2-u_1}\right|$ For small objects only: $m_L=-\frac{dv}{du}$ $m_L=-\frac{dv}{du}={\left[\frac{v}{u}\right]}^2=m^2$
Superficial magnification	$m_s=\frac{\text{Area of Image}}{\text{Area of Object}}=\frac{(ma)\times (mb)}{(a\times b)}=m^2$
Volume Magnification	$m_v=\frac{\text{Volume of Image}}{\text{Volume of Object}}=\frac{(ma)\times (mb)}{(a\times b}=m^4$
Velocity component along axis (Longitudinal velocity)	${\vec{v}}_x=-\frac{v^2}{u^2}{\vec{v}}_{ax}=-m^2{\vec{v}}_{ax}$
Velocity component perpendicular to  axis (Transverse velocity)	${\vec{v}}_y=\left[\frac{{\vec{v}}_{oy}}{m}+\frac{m{h}_o^2}{f}{\vec{v}}_{ox}\right]$
Newton's Formula	$XY=f^2$
Relative refractive index	${\mu }_1{u}_2={\mu }_2{u}_1\Rightarrow \left(\frac{{\mu }_2}{{\mu }_1}\right)=\left(\frac{u_1}{u_2}\right)=\frac{v_1}{v_2}$
Laws of Refraction	$\frac{\sin i}{\sin r}=\text{Constant}\Rightarrow \text{Snell’s Law}\Rightarrow \frac{\sin i}{\sin r}=\frac{n_2}{n_1}=\frac{v_1}{v_2}=\frac{{\lambda }_1}{{\lambda }_2}$ $n_1\sin i=n_2\sin r$
Apparent Depth	$d^{\prime }=\frac{d}{(\frac{n_r}{n_i})}$ $n_{\text{rel}}=\frac{n_i}{n_r}$ $d^{\prime }=\frac{d}{n_{\text{rel}}}\text{and}$$v^{\prime }=\frac{v}{n_{\text{relative}}}$
Critical Angle and Total Internal Reflection ( T. I. R.)	${\theta }_c={\sin }^{-1}\left(\frac{n_r}{n_d}\right)$
Angle of Deviation ()	$\delta =i+e-A$
Lens-Maker’s formula	$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Simple microscope	$m_0={\left(1+\frac{D}{f}\right)}_{\text{max}}\text{and}{m}_0={\left(\frac{D}{f}\right)}_{\text{min}}$
Compound microscope	$m_D=-\left(\frac{v_0-f_0}{f_0}\right)\left(1+\frac{D}{f_e}\right)$ $m_{\infty }=-\left(\frac{v_0-f_0}{f_0}\right)\cdot \frac{D}{f_e}$
Astronomical Telescope	$m_D=\frac{f_0}{f_e}\left(1+\frac{f_e}{D}\right)\text{and}{m}_{\infty }=-\frac{f_0}{f_e}$
Resultant Intensity	$I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos \Delta \varphi$
Fringe width()	$\beta =\frac{\lambda D}{d}$
Fresnel’s Distance (Zf)	$Z_f=\frac{a^2}{\lambda }$
Malus law	$I=I_0{\cos }^2\theta$
Brewster’s Law	$\tan {\theta }_{\rho p}=\frac{{\mu }_2}{{\mu }_1}$
Bright fringes is formed(Interference)	$y_{\text{max nth}}=\frac{n\lambda D}{d}$
Dark fringes is formed(Interference)	$y_{\text{min nth}}=\frac{(2n-1)\lambda D}{2d}$
Path Difference(Diffraction)	$\Delta x=a\sin \theta$
nth Secondary Minima (Diffraction)	$y=\frac{n\lambda D}{A}$
nth secondary Maxima	$y=\frac{(2n+1)\lambda D}{2a}$

3.0JEE Main Past Year Questions with Solutions on Optics

Q-1.If the refractive index of the material of a prism is $\cot \frac{A}{2}$ where A is the angle of prism then the angle of minimum deviation will be

$(1)\pi -2A(2)\frac{\pi }{2}-2A(3)\pi -A(4)\frac{\pi }{2}-A$

Solution: Ans(1)

$\cot \frac{A}{2}=\frac{\sin \left(\frac{A+{\delta }_{\min }}{2}\right)}{\sin \frac{A}{2}}$

$\Rightarrow \cos \frac{A}{2}=\sin \left(\frac{A+{\delta }_{\min }}{2}\right)$

$\Rightarrow \frac{A+{\delta }_{\min }}{2}=\frac{\pi }{2}-\frac{A}{2}$

${\delta }_{\min }=\pi -2A$

Q-2.Two immiscible liquids of refractive indices $\frac{8}{5}$ and $\frac{3}{2}$ respectively are put in a beaker as shown in the figure. The height of each column is 6 cm. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\frac{\alpha }{4}$ The value of $\alpha$ is______.

Solution:

Ans (31 cm )

$h_{\text{app}}=\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}=\frac{6}{2}+\frac{6}{\frac{8}{5}}=4+\frac{15}{4}=\frac{31}{4}\text{ cm}$

Q-3.A convex mirror of radius of curvature 30 cm forms an image that is half the size of the object. The object distance is :

(1) –15 cm        

(2) 45 cm            

(3) –45cm             

(4) 15 cm

Solution:

Ans (1)

R=30 cm

$f=\frac{R}{2}=+15cm$

Magnification $m=\mp \frac{1}{2}$

For convex mirror,virtual image is formed for real object, therefore,m is +ve

$\frac{1}{2}=\frac{f}{f-u}\to u=-15cm$

Q-4.A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

(1) –16 cm        

(2) -160 cm            

(3) +160cm             

(4) +16 cm

Solution:

Ans(2)

${\mu }_l=1.5,{\mu }_m=1.6,f_a=20\text{cm}$

$\frac{f_m}{f_a}=\frac{({\mu }_l-1){\mu }_m}{({\mu }_l-{\mu }_m)}$

$\frac{f_m}{20}=\frac{(1.5-1)1.6}{(1.5-1.6)}\Rightarrow f_m=-160\text{cm}$

Q-5.In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The y-x plot is shown in figure.The focal length of the lens is_____cm.

Solution:

Ans (20 cm)

$\frac{1}{f+20}-\frac{1}{-(f+20)}=\frac{1}{f}$

$\frac{2}{f+20}=1f\to f=20cm$

Or

$x_1{x}_2=f^2$ gives f=20 cm 

Q-6.The distance between the object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is ______ cm.

Solution:

Ans(+10 cm)

$\frac{v}{u}=-2\to v=-2u.......(1)$

v-u=45..........(2)

u=-15 cm

v=30 cm

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\to \frac{1}{f}=\frac{1}{30}\frac{-1}{-15}\to f=+10cm$

Q-7.If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be :

(1) 15 cm        

(2) -12 cm            

(3) -10cm             

(4) 103 cm

Solution:

Ans (3)

$\frac{-v}{u}$

$2=\frac{-(15-u)}{-u}$

$2u=15-u\to 3u=15u=5cm$

$\check{=}15-u\to 15-5=10cm$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{f}=\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10}$

f=-10 cm

Q-8.The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is __________ cm.

Solution:

Ans (15 cm)

v=3u

v-u=20 cm

$2u=20cm\to u=10cm$

$\frac{1}{f}=\frac{1}{(-30)}-\frac{1}{(-10)}\to f=15cm$

Q-9.Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at____ cm from the object. 

Solution:

Ans (200)

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\frac{1.5}{v}-\frac{1}{-100}=\frac{1.5-1}{20}$

Distance from object =100+100=200 cm

Q-10.The refractive index of prism is  and the ratio of the angle of minimum deviation to the angle of prism is one. The value of the angle of the prism is _________°.  

Solution:

Ans (60°) 

${\delta }_{\min }$

$r_1=r_2=\frac{A}{2}$

$\frac{{\delta }_{\min }}{A}=1$

$\frac{2i-A}{A}=1\to 2i=2A\to i=A$

Snell’s Law

1 ✕ sini= $\mu \sin r$

$\sin i=\mu \sin \left(\frac{A}{2}\right)$

$2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)=\sqrt{3}\sin \left(\frac{A}{2}\right)$

$\cos \left(\frac{A}{2}\right)=\frac{\sqrt{3}}{2}$

$\frac{A}{2}=30^{\circ }\Rightarrow A=60^{\circ }$

Q-11.A light ray is incident on a glass slab of thickness  cm and refractive index . The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of the ray after passing through the glass slab is ____cm. (Given sin 15° = 0.25)

Solution:

Ans (2 cm)

$i={\theta }_c$

$i={\sin }^{-1}\left(\frac{1}{\mu }\right)$

i=45° 

According to Snell’s Law

$1\sin 45^{\circ }=\sqrt{2}\sin r\Rightarrow r=30^{\circ }$

$\text{Lateral displacement}\Delta =\frac{t\sin (i-r)}{\cos r}$

$\Delta =\frac{4\sqrt{3}\times \sin 15^{\circ }}{\cos 30^{\circ }}\Rightarrow \Delta =2\text{cm}$

Q-12.A parallel beam of monochromatic light of wavelength 5000 Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on the screen, placed on its focal plane.  The first minima will be formed for the angle of diffraction of ________ (degree).

Solution:

Ans (30°)

$\text{For first Minima},a\sin \theta =n\lambda =1\lambda$

$\sin \theta =\frac{\lambda }{a}=\frac{5000\times 10^{-10}}{0.001\times 10^{-3}}=0.5$

$\theta =30^{\circ }$

Q-13.In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D = 0.2 m. the minimum distance between the slits S₁ and S₂ is  ______ mm.

Solution:

Ans (0.20 mm)

Path difference for minima at P

$2\sqrt{D^2+d^2}-2d=\frac{\lambda }{2}$

$\sqrt{D^2+d^2}-d=\frac{\lambda }{4}$

$\sqrt{D^2+d^2}=\frac{\lambda }{4}+d$

$D^2+d^2=D^2+\frac{{\lambda }^2}{16}+\frac{D\lambda }{2}$

$d^2=\frac{D\lambda }{2}+\frac{{\lambda }^2}{16}$

$d^2=\frac{0.2\times 400\times 10^{-9}}{2}+\frac{4\times 10^{-14}}{4}$

$d^2=400\times 10^{-10}$

$d=20\times 10^{-5}$

d= 0.20 mm

Q-14.The diffraction pattern of a light of wavelength 400 nm diffracting from a slit of width 0.2 mm is focused on the focal plane of a convex lens of focal length 100 cm. The width of the 1^st secondary maxima will be :
(1)2 mm          

(2) 2 cm          

(3) 0.02 mm               

(4) 0.2 mm

Solution:

Ans (1)

Width of 1^st secondary maxima = $\frac{\lambda D}{a}=\frac{400\times 10^{-9}\times 1}{0.2\times 10^{-3}}=\frac{2\times 10^{-7}}{10^{-4}}$

$2\times 10^{-}3m=2mm$             

Q-15.In Young’s double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is $\frac{7\pi }{4}$. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :

(1) 12         

(2) 34         

(3) 13            

4) 14

Solution:

Ans (1)

$\Delta x=\frac{7\pi }{4}$

$\Phi =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }\times \frac{7\pi }{4}=\frac{7\pi }{2}$

$I=I_{\text{max}}{\cos }^2\left(\frac{\Phi }{2}\right)$

$\frac{I}{I_{\text{max}}}={\cos }^2\left(\frac{\Phi }{2}\right)={\cos }^2\left(\frac{7\pi }{2\times 2}\right)={\cos }^2\left(\frac{7\pi }{4}\right)={\cos }^2\left(2\pi -\frac{7\pi }{4}\right)={\cos }^2\left(\frac{\pi }{4}\right)$

Q-16.A monochromatic light of wavelength 6000Å is incident on the single slit of width 0.01 mm. If the diffraction pattern is formed at the focus of the convex lens of focal length 20 cm, the linear width of the central maximum is :

(1) 60 mm       

(2) 24 mm        

(3) 120 mm          

(4) 12 mm

Solution:

Ans (2)

$\text{Linear Width}=\frac{2nd}{a}=\frac{2\times 6\times 10^{-7}\times 0.2}{1\times 10^{-5}}$

$2.4\times 10^{-2}=24mm$

Q-17.In Young's double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are 1.0 mm apart and the screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is___×10^–6 m.

Solution:

Ans (125)

Let intensity of light on screen due to each slit is I₀  

So intensity at centre of screen is 4I₀

Intensity at distance y from centre-

$I=I_0+I_0+2\sqrt{I_0{I}_0}\cos \Phi$

$I_{\text{max}}=4I_0$

$\frac{I_{\text{max}}}{2}=2I_0=2I_0+2I_0\cos \Phi$

$\cos \Phi =0\Rightarrow \Phi =\frac{\pi }{2}$

$k\Delta x=\frac{\pi }{2}$

$\frac{2\pi }{\lambda }d\sin \theta =\frac{\pi }{2}$

$\frac{2}{\lambda }d\times \frac{y}{D}=\frac{1}{2}$

$y=\frac{\lambda D}{4d}=\frac{5\times 10^{-7}\times 1}{5\times 10^{-3}}=125\times 10^{-6}$

Q-18.A microwave of wavelength 2.0 cm falls normally on a slit of width 4.0 cm. The angular spread of the central maxima of the diffraction pattern obtained on a screen 1.5 m away from the slit, will be:

$$\begin{gathered} (1)30\circ \\ (2)15\circ \\ (3)60\circ \\ (4)45\circ \end{gathered}$$

Solution:

Ans (3)

$\text{For first minima }a\sin \theta =\lambda$

$\sin \theta =\frac{\lambda }{a}=\frac{1}{2}\Rightarrow \theta =60^{\circ }$

Angular Spread=30°

Q-19.In a single slit diffraction pattern, a light of wavelength 6000 Å is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from slits. The width of the slit is ______ × 10^–4 m.

Solution:

Ans. (2)

For $n^{th}$ minima

$b\sin \theta =n\lambda$

$\lambda \text{ is so small so}\sin \theta \text{ is small, hence}\sin \theta \approx \tan \theta$

$b\tan \theta =n\lambda$

$b\frac{y}{D}=n\lambda \Rightarrow y_n=\frac{n\lambda D}{b}\text{position of }{n}^{\text{th}}\text{ minima}$

$B\to 1^{\text{st}}minima,A\to 3^{\text{rd}}minima$

$y_3=\frac{3\lambda D}{b},y_1=\frac{\lambda D}{b}$

$\Delta y=y_3-y_1=\frac{2\lambda D}{b}$

$3\times 10^{-3}=\frac{2\times 6000\times 10^{-10}\times 0.5}{b}$

$b=\frac{2\times 6000\times 10^{-10}\times 0.5}{3\times 10^{-3}}=2\times 10^{-4}\Rightarrow b=2$

Width of the slit is $2\times 10^{-4}m$

Q-20.When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :

$$\begin{gathered} (1)60\circ \\ (2)30\circ \\ (3)90\circ \\ (4)45\circ \end{gathered}$$

Solution:

Ans(4)

Let IO be the intensity of the unpolarised light incident on the first polaroid.

$I_1=\text{Intensity of light transmitted from 1 superscriptst polaroid}=\frac{I_0}{2}$

$\theta \text{be the angle between}{1}^{\text{st}}and{2}^{\text{nd}}polaroid$

$\theta +\varphi =90^{\circ }(as{1}^{st}and{3}^{rd}\text{polaroid are crossed})$

$\varphi =90^{\circ }-\theta$

$I_2=\text{Intensity from}{2}^{nd}\text{polaroid}$

$I_2=I_1{\cos }^2\theta =\frac{I_0}{2}{\cos }^2\theta$

$I_3=\text{Intensity from}{3}^{rd}\text{polaroid}$

$I_3=I_2{\cos }^2\varphi$

$I_3=I_1{\cos }^2\theta {\cos }^2\varphi$

$I_3=\frac{I_0}{2}{\cos }^2\theta {\cos }^2\varphi$

$\varphi =90^{\circ }-\theta$

$I_3=\frac{I_0}{2}{\cos }^2\theta {\sin }^2\theta$

$I_3=\frac{I_0}{2}{\left[\frac{2\sin \theta \cos \theta }{2}\right]}^2$

$I_3=\frac{I_0}{8}{\sin }^{2}2\theta$

$I_3\text{ will be maximum when }\sin 2\theta =1$

$2\theta =90^{\circ }\Rightarrow \theta =45^{\circ }$