A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is ____ mm.
Numerical Answer Type:
Enter a numerical valueAnswer: 280 to 280+4 marks
Solution & Explanation
### Related Formula
textMagnification m = fracD_textoutD_textin = fracf_2f_1$$\text{Magnification } m = \frac{D_{\text{out}}}{D_{\text{in}}} = \frac{f_2}{f_1}$$
### Core Logic
For a beam expander using two convex lenses, the lenses are arranged such that their focal points coincide. This creates an afocal system where parallel input rays remain parallel upon output.
Lenses solution diagram for Q46 - JEE Main 2026 Morning
By similar triangles at the focal point:
tan theta = fracD_1 / 2f_1 = fracD_2 / 2f_2$$\tan \theta = \frac{D_1 / 2}{f_1} = \frac{D_2 / 2}{f_2}$$
### Step 1: Solving for f2
fracD_1f_1 = fracD_2f_2$$\frac{D_1}{f_1} = \frac{D_2}{f_2}$$frac240 = frac14f_2$$\frac{2}{40} = \frac{14}{f_2}$$f_2 = 14 times frac402 = 14 times 20 = 280text mm$$f_2 = 14 \times \frac{40}{2} = 14 \times 20 = 280\text{ mm}$$
### Pattern Recognition
A standard beam expander (Keplerian telescope design used in reverse) has f_2/f_1 = D_2/D_1$f_2/f_1 = D_2/D_1$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
More Ray Optics and Optical Instruments Previous-Year Questions
Q50jee_main_2026_21_jan_morningMicroscope
In a microscope the objective is having focal length f_0 = 2text cm$f_{0} = 2\text{ cm}$ and eye-piece is having focal length f_e = 4text cm$f_{e} = 4\text{ cm}$. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____.
Numerical Answer.Answer: 100 to 100
Solution
### Related Formula
m approx fracLf_0 times fracDf_e$$m \approx \frac{L}{f_0} \times \frac{D}{f_e}$$
### Core Logic
For a compound microscope in normal adjustment (image formed at infinity), the magnifying power is given by the standard approximation:
m simeq fracl Df_0 f_e$$m \simeq \frac{l D}{f_{0} f_{e}}$$
where l$l$ is the tube length, D = 25text cm$D = 25\text{ cm}$ is the least distance of distinct vision.
### Step 1: Substitute Values
Given values:
l = 32text cm$l = 32\text{ cm}$f_0 = 2text cm$f_0 = 2\text{ cm}$f_e = 4text cm$f_e = 4\text{ cm}$D = 25text cm$D = 25\text{ cm}$ (standard assumption when not given)
m = frac322 times frac254$$m = \frac{32}{2} \times \frac{25}{4}$$m = 16 times frac254 = 4 times 25 = 100$$m = 16 \times \frac{25}{4} = 4 \times 25 = 100$$Microscope solution diagram for Q50 - JEE Main 2026 Morning
### Pattern Recognition
Compound microscope formulas: Normal adjustment to$\to$ image at infty$\infty$, m = (L/f_0)(D/f_e)$m = (L/f_0)(D/f_e)$. Image at near point D$D$, m = (L/f_0)(1 + D/f_e)$m = (L/f_0)(1 + D/f_e)$. Default to standard approximation when given tube length.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Qjee_main_2025_02_april_eveningSpherical Mirrors and Magnification
Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 \, textcm$12 \, \text{cm}$ , at the same distance of 18 \, textcm$18 \, \text{cm}$ from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is:
A.1 / 2$1 / 2$
B.2$2$
C.3$3$
D.1 / 3$1 / 3$
Solution
### Related Formula
1. Mirror focal length:
f = fracR2$$f = \frac{R}{2}$$
2. Magnification (image size relative to object size):
m = frach_ih_o = fracff - u$$m = \frac{h_i}{h_o} = \frac{f}{f - u}$$
where u$u$ is the object distance.
### Core Logic
Given parameters:
- Radius of curvature R = 12 \ mathrmcm implies |f| = 6 \ mathrmcm$R = 12 \ \mathrm{cm} \implies |f| = 6 \ \mathrm{cm}$
- Object distance u = -18 \ mathrmcm$u = -18 \ \mathrm{cm}$
Let's calculate magnification for both mirrors:
1. **For Convex Mirror:**
- Focal length f_textconvex = +6 \ mathrmcm$f_{\text{convex}} = +6 \ \mathrm{cm}$ (using Cartesian sign convention)
- Magnification:
m_1 = fracf_textconvexf_textconvex - u = frac66 - (-18) = frac624 = frac14$$m_1 = \frac{f_{\text{convex}}}{f_{\text{convex}} - u} = \frac{6}{6 - (-18)} = \frac{6}{24} = \frac{1}{4}$$
Thus, image size is frac14 h_o$\frac{1}{4} h_o$.
### Step 1: Calculate magnification of concave mirror
2. **For Concave Mirror:**
- Focal length f_textconcave = -6 \ mathrmcm$f_{\text{concave}} = -6 \ \mathrm{cm}$
- Magnification:
Convex and concave mirror ray diagram representationsm_2 = fracf_textconcavef_textconcave - u = frac-6-6 - (-18) = frac-612 = -frac12$$m_2 = \frac{f_{\text{concave}}}{f_{\text{concave}} - u} = \frac{-6}{-6 - (-18)} = \frac{-6}{12} = -\frac{1}{2}$$
Thus, image size is frac12 h_o$\frac{1}{2} h_o$.
### Step 2: Calculate the ratio of image sizes
Since the objects are identical (same height h_o$h_o$), the ratio of the sizes of the images is:
Convex and concave mirror ray diagram representationstextRatio = frac|h_i1||h_i2| = frac|m_1||m_2| = frac1/41/2 = frac12$$\text{Ratio} = \frac{|h_{i1}|}{|h_{i2}|} = \frac{|m_1|}{|m_2|} = \frac{1/4}{1/2} = \frac{1}{2}$$
Thus, the ratio of sizes is 1/2$1/2$.
### Pattern Recognition
Sees: Parallel convex vs concave mirror magnification.
Trap: Reversing the signs of focal lengths (Convex focal length is +$+$, Concave is -$-$ in standard coordinate systems).
Shortcut: Use direct magnification equation m = fracff-u$m = \frac{f}{f-u}$. For convex, m = frac624 = frac14$m = \frac{6}{24} = \frac{1}{4}$. For concave, m = frac-612 = -frac12$m = \frac{-6}{12} = -\frac{1}{2}$. Ratio of absolute values is frac1/41/2 = 1/2$\frac{1/4}{1/2} = 1/2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q17jee_main_2025_02_april_eveningLens Maker's Formula
A bi-convex lens has radius of curvature of both the surfaces same as 1/6$1/6$ cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides ( R_1 neq R_2$R_1 \neq R_2$ ), without any change in lens power then possible combination of R_1$R_1$ and R_2$R_2$ is:
A.frac13 mathrm~cm$\frac{1}{3} \mathrm{~cm}$ and frac13 mathrm~cm$\frac{1}{3} \mathrm{~cm}$
B.frac15 mathrm~cm$\frac{1}{5} \mathrm{~cm}$ and frac17 mathrm~cm$\frac{1}{7} \mathrm{~cm}$
C.frac13 mathrm~cm$\frac{1}{3} \mathrm{~cm}$ and frac17 mathrm~cm$\frac{1}{7} \mathrm{~cm}$
D.frac16 mathrm~cm$\frac{1}{6} \mathrm{~cm}$ and frac19 mathrm~cm$\frac{1}{9} \mathrm{~cm}$
Solution
### Related Formula
Lens Maker's Formula:
frac1f = (mu - 1) left(frac1R_1 - frac1R_2right)$$\frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
For a bi-convex lens of equal radii (R_1 = +R$R_1 = +R$, R_2 = -R$R_2 = -R$):
frac1f = (mu - 1) frac2R$$\frac{1}{f} = (\mu - 1) \frac{2}{R}$$
Power P propto frac1f$P \propto \frac{1}{f}$.
### Core Logic
For the initial lens:
- R = frac16 \ mathrmcm$R = \frac{1}{6} \ \mathrm{cm}$
- frac1f_1 = (mu - 1) frac21/6 = 12 (mu - 1)$\frac{1}{f_1} = (\mu - 1) \frac{2}{1/6} = 12 (\mu - 1)$
For the replacement lens (R_1 = +R_1$R_1 = +R_1$, R_2 = -R_2$R_2 = -R_2$):
frac1f_2 = (mu - 1) left(frac1R_1 + frac1R_2right)$$\frac{1}{f_2} = (\mu - 1) \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$
Since power must be preserved (f_1 = f_2$f_1 = f_2$):
frac1R_1 + frac1R_2 = frac2R = 12 \ mathrmcm^-1$$\frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} = 12 \ \mathrm{cm}^{-1}$$
We need to find a combination where the sum of the reciprocals of the radii equals 12$12$.
### Step 1: Verify the options
Let's check each choice:
- **Option (1):** R_1 = frac13$R_1 = \frac{1}{3}$, R_2 = frac13$R_2 = \frac{1}{3}$:
frac11/3 + frac11/3 = 3 + 3 = 6 neq 12$$\frac{1}{1/3} + \frac{1}{1/3} = 3 + 3 = 6 \neq 12$$
- **Option (2):** R_1 = frac15$R_1 = \frac{1}{5}$, R_2 = frac17$R_2 = \frac{1}{7}$:
frac11/5 + frac11/7 = 5 + 7 = 12 quad checkmark quad text(Correct Combination)$$\frac{1}{1/5} + \frac{1}{1/7} = 5 + 7 = 12 \quad \checkmark \quad \text{(Correct Combination)}$$
- **Option (3):** R_1 = frac13$R_1 = \frac{1}{3}$, R_2 = frac17$R_2 = \frac{1}{7}$:
frac11/3 + frac11/7 = 3 + 7 = 10 neq 12$$\frac{1}{1/3} + \frac{1}{1/7} = 3 + 7 = 10 \neq 12$$
- **Option (4):** R_1 = frac16$R_1 = \frac{1}{6}$, R_2 = frac19$R_2 = \frac{1}{9}$:
frac11/6 + frac11/9 = 6 + 9 = 15 neq 12$$\frac{1}{1/6} + \frac{1}{1/9} = 6 + 9 = 15 \neq 12$$
Thus, only Option (2) meets the physical conditions.
### Pattern Recognition
Sees: Equivalent thin lens power with modified surfaces.
Trap: Neglecting sign convention for the second spherical surface during substitution.
Shortcut: If the radii are of the form 1/n$1/n$, then the sum of n_1 + n_2$n_1 + n_2$ must equal 2 times n_textinitial$2 \times n_{\text{initial}}$. Since initial n = 6$n = 6$, 2 times 6 = 12$2 \times 6 = 12$. The only pairing whose denominators add up to 12 is 5 + 7$5 + 7$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q23jee_main_2025_02_april_eveningPrism Formula and Minimum Deviation
A ray of light suffers minimum deviation when incident on a prism having angle of the prism equal to 60^circ$60^{circ}$ . The refractive index of the prism material is sqrt2$\sqrt{2}$ . The angle of incidence (in degrees) is ______.
Numerical Answer.Answer: 45 to 45
Solution
### Related Formula
1. Prism Formula relating refractive index to minimum deviation:
mu = fracsinleft(fracA + delta_m2right)sinleft(fracA2right)$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
2. Under the condition of minimum deviation:
i = fracA + delta_m2$$i = \frac{A + \delta_m}{2}$$
Thus, the formula simplifies to:
mu = fracsin isin(A/2)$$\mu = \frac{\sin i}{\sin(A/2)}$$
### Core Logic
Given parameters:
- Angle of prism A = 60^circ$A = 60^\circ$
- Refractive index of material mu = sqrt2$\mu = \sqrt{2}$
Substitute the parameters into the simplified formula:
sqrt2 = fracsin isin(60^circ / 2)$$\sqrt{2} = \frac{\sin i}{\sin(60^\circ / 2)}$$sqrt2 = fracsin isin(30^circ)$$\sqrt{2} = \frac{\sin i}{\sin(30^\circ)}$$
### Step 1: Solve for angle of incidence
Since sin(30^circ) = 0.5$\sin(30^\circ) = 0.5$:
sin i = sqrt2 times frac12 = frac1sqrt2$$\sin i = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$$
Solving for i$i$:
i = 45^circ$i = 45^\circ$
Thus, the angle of incidence is 45$45$ degrees.
### Pattern Recognition
Sees: Prism minimum deviation and refracting index relations.
Trap: Mistaking the computed angle of incidence (i$i$) for the angle of minimum deviation (delta_m$\delta_m$).
Shortcut: Snell's law at the symmetrical boundary reduces to mu = sin i / sin(A/2)$\mu = \sin i / \sin(A/2)$. Because A = 60^circ$A = 60^\circ$, the denominator is sin 30^circ = 1/2$\sin 30^\circ = 1/2$. Therefore, sin i = 0.5 times sqrt2 = 1/sqrt2 implies i = 45^circ$\sin i = 0.5 \times \sqrt{2} = 1/\sqrt{2} \implies i = 45^\circ$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
More Ray Optics and Optical Instruments Questions — jee_main_2026_21_jan_morning
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