Optics is one of the topics with the highest weightage in JEE Main exam. Students can expect 5-6 questions from Optics and its subtopics in JEE Main 2021 paper. Over the years there have been about 3-4 questions from Optics carrying a weightage of 12-16 marks.

Optics is part of both Part A (theory) and Part B (practical component) of JEE Main Physics Syllabus. Important topics from Optics include: 

• For Part A, some of the important topics from Ray Optics include Reflection and Refraction of light at plane and spherical surfaces, Deviation and Dispersion of light by a prism, Lens Formula, Combination of lenses; Wave optics include Wavefront and Huygens’ principle, Interference, Young’s double-slit experiment, Diffraction due to a single slit, Polarisation, Brewster’s law, uses of plane-polarized light and Polaroids, etc. 
• For Part B, important topics covered from Optics include focal length of a Convex mirror, Concave mirror, and Convex lens using the parallax method, plot of the angle of deviation vs angle of incidence for a triangular prism and Refractive index of a glass slab using a traveling microscope. 

Physics is considered moderate to difficult in terms of levels of difficulty, However, as compared to other sections, it is the toughest section in JEE Main Question Paper. Questions asked are usually conceptual and application-based. Optics is one of the scoring topics in the physics section. 

To assist students with physics preparation for JEE Main 2021, Collegedunia has accumulated the JEE Main Test Series on Optics according to the previous year trends.

JEE Main Optics Test Series: Previous Year Questions and Solutions

Ques. A new type of transparent plastic has been developed. When light travels through it, it only travels at 2x107 m/s. What is the refractive index (n) of this new plastic? (Year 2005)

1. n = 0.67
2. n = 1.5
3. n = 6.67
4. n = 15

Solution: An object's refractive index is referred to as the ratio of the speed of light in vacuum to the speed of light in a medium. Thus to find the refractive index of our new plastic, we need to divide the speed of light in a vacuum (c), by the speed that light travels in the medium (v). Since we know that c = 3x108 m/s and v = 2x107 m/s therefore with help of formula n = c/v, the refractive index is 15 (Option D).

Ques. Which of the following is true about light? (Year 2007)

1. It is an electromagnetic wave
2. It does not propagates in vacuum
3. Its maximum speed is approximately 3x108 m/s
   1. I only
   2. I and II only
   3. I and III only
   4. III only
   5. I, II and III

Solution : Light is an electromagnetic wave that can propagate in vacuum with a maximum speed of approximately 3x108 m/s. Therefore the answer is Option E. I, II and III

Ques. The speed of light in a certain material is 50% of its speed in vacuum. What is the refractive index of this material? (Year 2010)

1. 1.5
2. 0.5
3. 6.0
4. 2.0
5. 1.5×108

Solution : Since, v = 50% c (c speed of light in vacuum)

Definition of refractive index (n) is n = c / v = c / 50% c = 2.0

Therefore the answer is Option D. 2.0

Ques. Which of the following is true about light with a single wavelength? (Year 2017)

1. It can be refracted
2. It cannot be dispersed
3. It can be reflected
   1. I, II and III
   2. I and II only
   3. II and III only
   4. I and III only
   5. None

Solution : Light of any wavelength can be reflected and refracted. Light with a single wavelength cannot be dispersed. Therefore the answer is Option A.

Ques. What is the critical angle ic at the interface glass-cladding of an optical fiber whose core has a refractive index equal to 1.5 and cladding with a refractive index 1.45? (Year 2011)

1. 15°
2. 105°
3. 86°
4. 83°
5. 75°

Solution : Since ic = arcsin(1.45/1.5) = 75°, therefore the answer is Option E

Ques. A ray of light is incident in medium (1) where light has speed s1 onto the interface with medium (2) where light has speed s2. If s1 > s2, then at the interface? (Year 2018)

1. the ray will refract away from the normal to the interface
2. the ray will refract toward from the normal to the interface
3. the ray will follow a straight path from medium (1) to medium (2)
4. the angle of reflection is greater than the angle of incidence
5. the angle of reflection is not equal to the angle of incidence

Solution : As per Snell's law, n1 sin (i) = n2 sin ®

Use definition of refractive index: n = c / s (c speed of light in vacuum and s speed of light in the material)

(c/s1) sin (i) = (c / s2) sin (r)

sin (r) = (s2 / s1) sin (i)

s2 / s1 <1 leads to

sin(r) < sin (i)

r < i

The ray refract toward the normal. Therefore the answer is Option B

Ques. An object of height 10 cm is placed 50 cm in front of a bi-convex lens with a focal length of 20 cm. Which of the following is true about the image? (Year 2019)

1. The image is virtual
2. The image is situated on the opposite side as the object
3. The image is inverted
   1. I only
   2. I and II only
   3. II and III only
   4. II only
   5. III only

Solution : Let do be the distance from lens to object, di distance from lens to the image and f be the focal length. Use the lens equation

1/do + 1/di = 1/f

1/50 + 1/di = 1 / 20

1/di = 1/20 - 1/50

di = 33 cm

So di which is the distance from the lens to the image is positive, therefore the image is real and inverted and on the opposite side of the object. Therefore the answer is Option C

Ques. An object of height 5 cm is placed 25 cm in front of a bi-convex lens with a focal length of 10 cm. What is the height of the image? (Year 2006)

1. 2.5 cm
2. 12.5 cm
3. 6.8 cm
4. 3.4 cm
5. 7.4 cm

Solution : Let do be the distance from lens to object, di distance from lens to the image and f be the focal length.

Let hi be the height of the image and ho be the height of the object. First use the lens equation to find di.

1/do + 1/di = 1/f

1/25 + 1/di = 1/10

Solve to find di = 17 cm

Magnification equation gives

hi / ho = - di / do

hi / 5 = - 17 / 25

hi = - 3.4 cm

Therefore the answer is Option D

Ques. An optical fiber is made up of a core of refractive index 1.6 and a cladding of refractive index 1.5. What is the maximum angle that the light rays can make with the axis of the optical fiber so that light is totally reflected inside the optical fiber?

1. 10°
2. 15°
3. 20°
4. 70°
5. 90°

Solution : To have total internal reflection, the angle of incidence i must be larger than the critical angle ic.

i > ic

ic = arcsin(1.5/1.6) = 70 °

Let α be the angle made by the ray and the axis of the optical fiber. The relationship between α and i is:

α = 90° - i

i > ic

90° - α > 70°

α < 90° - 70°

α < 20 °

Therefore answer is Option C

Ques. For an object in front of a plane mirror, which of the following about its images is(are) true?

1. The image is real
2. The image is upright
3. The height of the image is twice the image of the object
   1. I, II and III
   2. I and II only
   3. II only
   4. I and III only
   5. None

Solution :The image of the object in front of a plane mirror is virtual, upright and of the same height as the object. Therefore the answer is Option C

Also Check JEE Main 2021 Sample Papers

JEE Main Optics: Some Important Formulas

Check JEE Main Notes on Wave Optics