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 10.
Q50 (2026)
In a microscope the objective is having focal length $f_{0} = 2\text{ cm}$ and eye-piece is having focal length $f_{e} = 4\text{ cm}$. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____.
### Related Formula
$$m \approx \frac{L}{f_0} \times \frac{D}{f_e}$$
### Core Logic
For a compound microscope in normal adjustment (image formed at infinity), the magnifying power is given by the standard approximation:
$$m \simeq \frac{l D}{f_{0} f_{e}}$$
where $l$ is the tube length, $D = 25\text{ cm}$ is the least distance of distinct vision.
### Step 1: Substitute Values
Given values:
$l = 32\text{ cm}$
$f_0 = 2\text{ cm}$
$f_e = 4\text{ cm}$
$D = 25\text{ cm}$ (standard assumption when not given)
$$m = \frac{32}{2} \times \frac{25}{4}$$
$$m = 16 \times \frac{25}{4} = 4 \times 25 = 100$$
{{SOL_IMG1}}
### Pattern Recognition
Compound microscope formulas: Normal adjustment $\to$ image at $\infty$, $m = (L/f_0)(D/f_e)$. Image at near point $D$, $m = (L/f_0)(1 + D/f_e)$. Default to standard approximation when given tube length.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q45 (2026)
As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.
{{IMG1}}
If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is.
- $60^{\circ}$
- $75^{\circ}$
- $90^{\circ}$
- $45^{\circ}$
### Related Formula
$A = r_1 + r_2$
$$\mu = \frac{\sin i}{\sin r}$$
### Core Logic
{{SOL_IMG1}}
Since the incident ray is parallel to the base, looking at the geometry, the angle of incidence on the first face can be derived. The base angle is $45^{\circ}$. Thus, the normal to the first surface makes an angle of $45^{\circ}$ with the incident ray. Therefore, $i_1 = 45^{\circ}$.
For the emergent ray to graze the second surface, the angle of emergence $e = 90^{\circ}$.
Applying Snell's Law at the second surface:
$$\mu \sin r_2 = 1 \cdot \sin 90^{\circ}$$
$$\sqrt{2} \sin r_2 = 1 \implies \sin r_2 = \frac{1}{\sqrt{2}}$$
$$r_2 = 45^{\circ}$$
### Step 1: First Surface Analysis
Applying Snell's Law at the incident surface:
$$1 \cdot \sin(45^{\circ}) = \mu \sin(r_1)$$
$$\frac{1}{\sqrt{2}} = \sqrt{2} \sin(r_1)$$
$$\sin(r_1) = \frac{1}{2} \implies r_1 = 30^{\circ}$$
### Step 2: Prism Geometry
The prism angle $A$ at the apex is the sum of the internal angles of refraction:
$$A = r_1 + r_2 = 30^{\circ} + 45^{\circ} = 75^{\circ}$$
### Step 3: Final Conclusion
In the triangle of the prism, the sum of angles is $180^{\circ}$.
$$A + 45^{\circ} + \theta = 180^{\circ}$$
$$75^{\circ} + 45^{\circ} + \theta = 180^{\circ}$$
$$120^{\circ} + \theta = 180^{\circ} \implies \theta = 60^{\circ}$$
### Pattern Recognition
Grazing emergence means $r_2 = \theta_c$ (critical angle). Horizontal incidence with a known base angle gives $i_1$. Combining these through $A = r_1 + r_2$ resolves the full geometry.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q39 (2026)
A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification $m_{1}$, when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to $m_{2}$. The value of $\left|\frac{m_{1}}{m_{2}}\right|$ is \_\_\_\_.
- $\frac{5}{9}$
- $\frac{5}{27}$
- $\frac{3}{2}$
- $\frac{25}{27}$
### Related Formula
$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}, \quad m = \frac{v}{u}$$
### Core Logic
This question was officially dropped by the examining authority. Full marks awarded to all candidates.
### Pattern Recognition
Sees: Lens combination with separation gap introduced.
Shortcut: Note official exam status (Dropped question).
Check: Question dropped in final key. ✓
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q40 (2026)
Consider an equilateral prism (refractive index $\sqrt{2}$). A ray of light is incident on its one surface at a certain angle i. If the emergent ray is found to graze along the other surface then the angle of refraction at the incident surface is close to \_\_\_\_.
- $15^{\circ}$
- $20^{\circ}$
- $40^{\circ}$
- $30^{\circ}$
### Related Formula
$$\mu \sin r_2 = 1 \sin e, \quad r_1 + r_2 = A$$
### Core Logic
{{SOL_IMG1}}
For grazing emergence ($e = 90^{\circ}$):
$$\mu \sin r_2 = 1 \implies \sqrt{2} \sin r_2 = 1 \implies r_2 = 45^{\circ}$$
Since prism is equilateral ($A = 60^{\circ}$):
$$r_1 = A - r_2 = 60^{\circ} - 45^{\circ} = 15^{\circ}$$
### Pattern Recognition
Sees: Equilateral prism with grazing emergence.
Shortcut: Use critical angle condition at second surface and prism angle relation $r_1 + r_2 = A$.
Check: Matches option (1). ✓
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q48 (2026)
A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm. Refractive index of glass is 1.5. The rays converge to a point at a distance x cm from the centre of the curvature of the spherical surface. The value of x is \_\_\_\_ cm.
### Related Formula
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$
### Core Logic
{{SOL_IMG1}}
For parallel beam incident from air ($u = \infty$):
$$\frac{1.5}{v} - \frac{1}{\infty} = \frac{1.5 - 1.0}{50} \implies v = 150 \text{ cm}$$
Distance from center of curvature:
$$x = v - R = 150 - 50 = 100 \text{ cm}$$
### Pattern Recognition
Sees: Refraction at convex spherical surface with parallel incident beam.
Shortcut: Apply spherical surface refraction formula with $u=\infty$.
Check: Numerical answer is 100 cm. ✓
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
