Ray Optics and Optical Instruments JEE Main PYQs (2020–2026)

Solved Ray Optics and Optical Instruments PYQs

3 recent questions with answers. The full set of 209 is free to practise.

1. JEE Main 2026EasyPrisms: Refraction, Dispersion, and Deviation
   Q1

   Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is $\_\_\_\_$

   • A1.5
   • B$\sqrt{3}$
   • C$\sqrt{2}$
   • D1.65

   Show answer & solution

   Correct answer: (C)

   For a prism with angle A and refractive index $\mu$ , the relationship between the angle of minimum deviation ( ${\delta }_m$ ) and refractive index is given by:
   
   $$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$
   
   For equilateral prism the angle of the prism is $A=60^{\circ }$ .
   
   The angle of minimum deviation is half the angle of prism $\left({\delta }_m=\frac{A}{2}\right)$ .
   
   Therefore, ${\delta }_m=\frac{60^{\circ }}{2}=30^{\circ }$
   
   Substituting $A=60^{\circ }$ and ${\delta }_m=30^{\circ }$ into the formula:
   
   $$\mu =\frac{\sin \left(\frac{60^{\circ }+30^{\circ }}{2}\right)}{\sin \left(\frac{60^{\circ }}{2}\right)}$$
   
   $$\Rightarrow \mu =\frac{\sin \left(45^{\circ }\right)}{\sin \left(30^{\circ }\right)}$$
   
   $$\Rightarrow \mu =\frac{1/\sqrt{2}}{1/2}=\frac{2}{\sqrt{2}}$$
   
   $$\Rightarrow \mu =\sqrt{2}$$
   
   Therefore, the refractive index of the prism is $\sqrt{2}$ .
   
   Thus the correct option is (C).
2. JEE Main 2026EasyThin Lenses: Lens Maker's Formula and Image Formation
   Q2

   A convex lens is made from glass material having refractive index of 1.4 with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is $\_\_\_\_$ .

   • A0.5
   • B2.5
   • C0.8
   • D1.25

   Show answer & solution

   Correct answer: (D)

   For a thin lens with two curved surfaces, the focal length ( $f$ ) is related to the refractive index ( $n$ ) and the radii of curvature $(R_1$ and $R_2)$ by Lens Makers’ formula,
   
   $$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
   
   The refractive index of the material, $n=1.4$ .
   
   The lens is convex with the same radius of curvature on both sides (R).
   
   According to the Cartesian sign convention:
   
   The first surface is convex towards the incident light, so $R_1=+R$ .
   
   The second surface is concave towards the incident light, so $R_2=-R$ .
   
   Substituting the values into the formula :
   
   $$\frac{1}{f}=(1.4-1)\left(\frac{1}{R}-\frac{1}{-R}\right)$$
   
   $$\Rightarrow \frac{1}{f}=(0.4)\left(\frac{1}{R}+\frac{1}{R}\right)$$
   
   $$\Rightarrow \frac{1}{f}=0.4\times \frac{2}{R}$$
   
   $$\Rightarrow \frac{1}{f}=\frac{0.8}{R}$$
   
   $$\Rightarrow f=\frac{R}{0.8}$$
   
   $$\Rightarrow \frac{f}{R}=\frac{1}{0.8}=1.25$$
   
   Therefore, the ratio of its focal length and radius of curvature is 1.25 .
   
   Thus, the correct option is (D).
3. JEE Main 2026EasyReflection at Plane and Spherical Mirrors
   Q3

   A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is
   
   $\_\_\_\_$ cm.

   Numerical answer type

   Show answer & solution

   Correct answer: 10

   For a mirror, magnification is given as $m=-\frac{v}{u}$ , where v is the image distance and u is the object distance.
   
   From the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ :
   
   $$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{uf}\Rightarrow v=\frac{uf}{u-f}$$
   
   Substituting v into the magnification formula:
   
   $$m=-\frac{f}{u-f}=\frac{f}{f-u}$$
   
   A concave mirror ( $f=-10cm$ ) produces a magnification $|m|=2$ in two cases:
   
   Case 1 : Real Image ( $m=-2$ )
   
   $$-2=-\frac{10}{-10-u_1}$$
   
   $$\Rightarrow 20+2u_1=-10\Rightarrow 2u_1=-30\Rightarrow u_1=-15cm$$
   
   Case 2 : Virtual Image ( $m=+2$ )
   
   $$2=-\frac{10}{-10-u_2}$$
   
   $$\Rightarrow -20-2u_2=-10\Rightarrow -2u_2=10\Rightarrow u_2=-5cm$$
   
   The distance between the two object positions is:
   
   $$\Delta u=\left|u_1-u_2\right|=|-15-(-5)|=|-10|=10cm$$
   
   Therefore, the distance between the two positions is 10 cm .

Topic-wise question count

Where JEE Main has asked from Ray Optics and Optical Instruments, most frequent first.

• Refraction at Plane Surfaces and Total Internal Reflection43
• Thin Lenses: Lens Maker's Formula and Image Formation41
• Reflection at Plane and Spherical Mirrors33
• Prisms: Refraction, Dispersion, and Deviation32
• Combination of Optical Elements and Silvered Lenses28
• Optical Instruments and Defects of Vision18
• Refraction at Spherical Interfaces14

Year-wise Ray Optics and Optical Instruments questions in JEE Main

Number of questions from this chapter in each year's paper(s).