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.
JEE Main Previous Year Solved Questions on Optics
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 aDλ≪a or a≫Dλ. If this condition is fulfilled, light is said to move rectilinearly.
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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°), 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° . 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
δ=360∘−2θ
Relation between velocity of object and image
For x-Axis
viG−vmG=−(voG−vmG)
for y axis and z axis
viG−vmG=(voG−vmG) or viG=voG
viG=velocity of image with respect to ground.
Number of images formed by two inclined mirrors
(1) θ360∘=even number;number of image=θ360∘−1
(2)If θ360∘=odd number;number of image=θ360∘−1 if the object is placed on the angle bisector.
(3) If θ360∘=odd number;number of image=θ360∘, if the object is not placed on the angle bisector.
(4) If 360°=Integerθ360∘=Integer, then count the number of images as explained above.
Mirror formula
v1+u1=f1
Lateral Magnification
m=hohi=−uv
Longitudinal magnification
mL=Length of ObjectLength of Image=u2−u1v2−v1
For small objects only:
mL=−dudv
mL=−dudv=[uv]2=m2
Superficial magnification
ms=Area of ObjectArea of Image=(a×b)(ma)×(mb)=m2
Volume Magnification
mv=Volume of ObjectVolume of Image=(a×b(ma)×(mb)=m4
Velocity component along axis (Longitudinal velocity)
vx=−u2v2vax=−m2vax
Velocity component perpendicular to axis (Transverse velocity)
Critical Angle and Total Internal Reflection ( T. I. R.)
θc=sin−1(ndnr)
Angle of Deviation ()
δ=i+e−A
Lens-Maker’s formula
f1=(μ−1)(R11−R21)
Simple microscope
m0=(1+fD)maxandm0=(fD)min
Compound microscope
mD=−(f0v0−f0)(1+feD)
m∞=−(f0v0−f0)⋅feD
Astronomical Telescope
mD=fef0(1+Dfe)andm∞=−fef0
Resultant Intensity
IR=I1+I2+2I1I2cosΔϕ
Fringe width()
β=dλD
Fresnel’s Distance (Zf)
Zf=λa2
Malus law
I=I0cos2θ
Brewster’s Law
tanθρp=μ1μ2
Bright fringes is formed(Interference)
ymax nth=dnλD
Dark fringes is formed(Interference)
ymin nth=2d(2n−1)λD
Path Difference(Diffraction)
Δx=asinθ
nth Secondary Minima (Diffraction)
y=AnλD
nth secondary Maxima
y=2a(2n+1)λD
3.0JEE Main Past Year Questions with Solutions on Optics
Q-1.If the refractive index of the material of a prism is cot2A where A is the angle of prism then the angle of minimum deviation will be
(1)π−2A(2)2π−2A(3)π−A(4)2π−A
Solution: Ans(1)
cot2A=sin2Asin(2A+δmin)
⇒cos2A=sin(2A+δmin)
⇒2A+δmin=2π−2A
δmin=π−2A
Q-2.Two immiscible liquids of refractive indices 58and 23 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 4α The value of α is______.
Solution:
Ans (31 cm )
happ=μ1h1+μ2h2=26+586=4+415=431 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=2R=+15cm
Magnification m=∓21
For convex mirror,virtual image is formed for real object, therefore,m is +ve
21=f−uf→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)
μl=1.5,μm=1.6,fa=20cm
fafm=(μl−μm)(μl−1)μm
20fm=(1.5−1.6)(1.5−1)1.6⇒fm=−160cm
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)
f+201−−(f+20)1=f1
f+202=1f→f=20cm
Or
x1x2=f2 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)
uv=−2→v=−2u.......(1)
v-u=45..........(2)
u=-15 cm
v=30 cm
f1=v1−u1→f1=301−15−1→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)
u−v
2=−u−(15−u)
2u=15−u→3u=15u=5cm
=ˇ15−u→15−5=10cm
f1=v1+u1
f1=101+(−5)1=101−2=10−1
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→u=10cm
f1=(−30)1−(−10)1→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)
vμ2−uμ1=Rμ2−μ1
v1.5−−1001=201.5−1
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°)
δmin
r1=r2=2A
Aδmin=1
A2i−A=1→2i=2A→i=A
Snell’s Law
1 ✕ sini= μsinr
sini=μsin(2A)
2sin(2A)cos(2A)=3sin(2A)
cos(2A)=23
2A=30∘⇒A=60∘
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=θc
i=sin−1(μ1)
i=45°
According to Snell’s Law
1sin45∘=2sinr⇒r=30∘
Lateral displacementΔ=cosrtsin(i−r)
Δ=cos30∘43×sin15∘⇒Δ=2cm
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°)
For first Minima,asinθ=nλ=1λ
sinθ=aλ=0.001×10−35000×10−10=0.5
θ=30∘
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 S1and S2 is ______ mm.
Solution:
Ans (0.20 mm)
Path difference for minima at P
2D2+d2−2d=2λ
D2+d2−d=4λ
D2+d2=4λ+d
D2+d2=D2+16λ2+2Dλ
d2=2Dλ+16λ2
d2=20.2×400×10−9+44×10−14
d2=400×10−10
d=20×10−5
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 1stsecondary maxima will be : (1)2 mm
(2) 2 cm
(3) 0.02 mm
(4) 0.2 mm
Solution:
Ans (1)
Width of 1st secondary maxima = aλD=0.2×10−3400×10−9×1=10−42×10−7
2×10−3m=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 47π. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :
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)
Linear Width=a2nd=1×10−52×6×10−7×0.2
2.4×10−2=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=I0+I0+2I0I0cosΦ
Imax=4I0
2Imax=2I0=2I0+2I0cosΦ
cosΦ=0⇒Φ=2π
kΔx=2π
λ2πdsinθ=2π
λ2d×Dy=21
y=4dλD=5×10−35×10−7×1=125×10−6
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:
(1)30∘(2)15∘(3)60∘(4)45∘
Solution:
Ans (3)
For first minima asinθ=λ
sinθ=aλ=21⇒θ=60∘
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 nth minima
bsinθ=nλ
λ is so small sosinθ is small, hencesinθ≈tanθ
btanθ=nλ
bDy=nλ⇒yn=bnλDposition of nth minima
B→1stminima,A→3rdminima
y3=b3λD,y1=bλD
Δy=y3−y1=b2λD
3×10−3=b2×6000×10−10×0.5
b=3×10−32×6000×10−10×0.5=2×10−4⇒b=2
Width of the slit is 2×10−4m
Q-20.When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :
(1)60∘(2)30∘(3)90∘(4)45∘
Solution:
Ans(4)
Let IO be the intensity of the unpolarised light incident on the first polaroid.
I1=Intensity of light transmitted from 1 superscriptst polaroid=2I0
θbe the angle between1stand2ndpolaroid
θ+ϕ=90∘(as1stand3rdpolaroid are crossed)
ϕ=90∘−θ
I2=Intensity from2ndpolaroid
I2=I1cos2θ=2I0cos2θ
I3=Intensity from3rdpolaroid
I3=I2cos2ϕ
I3=I1cos2θcos2ϕ
I3=2I0cos2θcos2ϕ
ϕ=90∘−θ
I3=2I0cos2θsin2θ
I3=2I0[22sinθcosθ]2
I3=8I0sin22θ
I3 will be maximum when sin2θ=1
2θ=90∘⇒θ=45∘
Table of Contents
1.0Introduction
2.0Key Concepts to Remember
2.1Fresnel’s Distance (Zf)
2.2Malus law
3.0JEE Main Past Year Questions with Solutions on Optics
