Ray Optics and Optical Instruments Previous Year Questions — JEE Main Physics

63 past-year Ray Optics and Optical Instruments questions from JEE Main (Physics) — page 13.

Q49 (2026)

A convex lens of refractive index 1.5 and focal length $f = 18 \mathrm{~cm}$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$ . The value of $\alpha$ is ____. (refractive index of water = 4/3)
### Related Formula $$\frac{1}{f_{\text{medium}}} = \left( \frac{\mu_L}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ ### Core Logic Use the Lens Maker's formula for the lens in air, and then for the lens in water. Divide the two equations to eliminate the radius terms and solve for the focal length in water. ### Step 1: Lens in Air $$\frac{1}{f_{\text{air}}} = \left( \frac{1.5}{1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ ### Step 2: Lens in Water $$\frac{1}{f_{\text{water}}} = \left( \frac{1.5}{4/3} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ $$= \left( \frac{4.5}{4} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ $$= \left( \frac{0.5}{4} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ ### Step 3: Ratio and Difference $$\frac{f_{\text{water}}}{f_{\text{air}}} = \frac{0.5}{0.5 / 4} = 4$$ $$f_{\text{water}} = 4f_{\text{air}} = 4f$$ Difference in focal length: $$f_{\text{water}} - f_{\text{air}} = 4f - f = 3f$$ Therefore, $\alpha = 3$. ### Pattern Recognition Standard result to memorize: A glass lens ($\mu = 1.5$) immersed in water ($\mu = 4/3$) has its focal length become exactly 4 times its focal length in air. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Q28 (2026)

A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. {{IMG1}} The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of $30 \text{ cm}$ from the biconvex lens, the magnification of the image will be :
  1. $-2$
  2. $+2$
  3. $+2.5$
  4. $-2.5$
### Related Formula $$\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\frac{1}{f_{\text{net}}} = \frac{1}{f_1} + \frac{1}{f_2}$$ $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{\text{net}}}$$ $$m = \frac{v}{u}$$ ### Core Logic The biconvex lens acts as a combination of two thin lenses in contact. For lens 1 (left): $\mu_1 = 1.5$, $R_{1} = 15 \text{ cm}$, flat side radius $R_2 = \infty$. For lens 2 (right): $\mu_2 = 1.2$, flat side radius $R_1 = \infty$, curved side radius $R_2 = -12 \text{ cm}$ (applying sign convention). ### Step 1: Calculate Focal Lengths $$\frac{1}{f_1} = (1.5 - 1) \left(\frac{1}{15} - \frac{1}{\infty}\right) = 0.5 \times \frac{1}{15} = \frac{1}{30}$$ $$\frac{1}{f_2} = (1.2 - 1) \left(\frac{1}{\infty} - \frac{1}{-12}\right) = 0.2 \times \frac{1}{12} = \frac{1}{60}$$ ### Step 2: Calculate Equivalent Focal Length $$\frac{1}{f_{\text{net}}} = \frac{1}{30} + \frac{1}{60} = \frac{2 + 1}{60} = \frac{3}{60} = \frac{1}{20}$$ So, $f_{\text{net}} = +20 \text{ cm}$. ### Step 3: Image Distance and Magnification Given object distance $u = -30 \text{ cm}$. $$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$$ $$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$$ $$v = +60 \text{ cm}$$ Magnification $m = \frac{v}{u} = \frac{60}{-30} = -2$. ### Pattern Recognition Combination of thin lenses in contact: simply add their powers $1/f_{net} = 1/f_1 + 1/f_2$. Apply lens maker's formula carefully with correct sign conventions for the planar boundary ($R=\infty$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

Q43 (2026)

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.
  1. $\sqrt{2} < n < 2\sqrt{2}$
  2. $1 < n < 2$
  3. $n \geq 2$
  4. $\sqrt{2} < n < 2$
### Related Formula $$\mu = \frac{\sin\left(\frac{\delta_{\min} + A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ $$\delta_{\min} = 2i - A$$ ### Core Logic Given $\delta_{\min} = A$. $$\mu = n = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin(A/2)}$$ $$n = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} = 2 \cos(A/2)$$ Since $A > 0$, $\cos(A/2) < 1$. Thus, $n < 2$. ### Step 1: Applying Physical Bounds At minimum deviation, $i = \frac{\delta_{\min} + A}{2}$. Since $\delta_{\min} = A$, we have $i = \frac{2A}{2} = A$. For light to enter the prism, the angle of incidence must be less than $90^\circ$ (grazing incidence is the limit). Thus, $i < 90^\circ \Rightarrow A < 90^\circ$. ### Step 2: Calculate Limits for n Since $A < 90^\circ$, we have $A/2 < 45^\circ$. The cosine function is strictly decreasing in the first quadrant, so: If $A/2 < 45^\circ$, then $\cos(A/2) > \cos(45^\circ) = \frac{1}{\sqrt{2}}$. Therefore: $$n = 2 \cos(A/2) > 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$$ Combining the limits: $$\sqrt{2} < n < 2$$ ### Pattern Recognition When $\delta_{min} = A$, $\mu = 2\cos(A/2)$. The extreme bounding conditions arise from $A > 0$ and $i < 90^\circ$. This guarantees the theoretical range $\sqrt{2} < \mu < 2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments
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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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