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 — Page 9
Q34jee_main_2024_29_jan_morningSpherical Mirrors
A convex mirror of radius of curvature 30 mathrm~cm$30 \mathrm{~cm}$ forms an image that is half the size of the object. The object distance is:
A.-15 mathrm~cm$-15 \mathrm{~cm}$
B.45 mathrm~cm$45 \mathrm{~cm}$
C.-45 mathrm~cm$-45 \mathrm{~cm}$
D.15 mathrm~cm$15 \mathrm{~cm}$
Solution
### Related Formula
The magnification (m$m$) of a spherical mirror is given by:
m = fracff - u$$m = \frac{f}{f - u}$$
where,
f$f$ = focal length (R/2$R/2$)
u$u$ = object distance
### Core Logic
Given radius of curvature, R = 30 mathrm~cm$R = 30 \mathrm{~cm}$.
For a convex mirror:
f = +fracR2 = +15 mathrm~cm$$f = +\frac{R}{2} = +15 \mathrm{~cm}$$
Since a convex mirror always forms a virtual, erect, and diminished image of a real object, the magnification m$m$ must be positive:
m = +frac12$$m = +\frac{1}{2}$$Ray diagram for convex mirror showing image formation for Q34 - JEE Main 2024 Morning
### Step 1: Solve for Object Distance
Using the magnification formula:
+frac12 = frac1515 - u$$+\frac{1}{2} = \frac{15}{15 - u}$$15 - u = 30 implies u = -15 mathrm~cm$$15 - u = 30 \implies u = -15 \mathrm{~cm}$$
Thus, the object distance is -15 mathrm~cm$-15 \mathrm{~cm}$.
### Pattern Recognition
Remember: Convex mirrors produce *only* virtual images for real objects, which means m$m$ is always positive and less than 1$1$. If the question mentioned a concave mirror with a diminished image of half size, the image would be real, and m$m$ would be negative.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q45jee_main_2024_29_jan_morningRefraction at Spherical Surfaces and by Lenses
A biconvex lens of refractive index 1.5 has a focal length of 20 mathrm~cm$20 \mathrm{~cm}$ in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:
A.-16 mathrm~cm$-16 \mathrm{~cm}$
B.-160 mathrm~cm$-160 \mathrm{~cm}$
C.+160 mathrm~cm$+160 \mathrm{~cm}$
D.+16 mathrm~cm$+16 \mathrm{~cm}$
Solution
### Related Formula
From the Lens Maker's Formula:
frac1f = left( fracmu_textlensmu_textmedium - 1 right) left( frac1R_1 - frac1R_2 right)$$\frac{1}{f} = \left( \frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$
Taking the ratio of focal length in liquid medium (f_m$f_m$) to focal length in air (f_a$f_a$):
fracf_mf_a = frac(mu_1 - 1) mu_mmu_1 - mu_m$$\frac{f_m}{f_a} = \frac{(\mu_1 - 1) \mu_m}{\mu_1 - \mu_m}$$
where,
mu_1$\mu_1$ = refractive index of the lens material = 1.5
mu_m$\mu_m$ = refractive index of the liquid medium = 1.6
f_a$f_a$ = focal length in air = 20 mathrm~cm$20 \mathrm{~cm}$
### Core Logic
Substitute the parameters into the relative ratio template equation:
fracf_m20 = frac(1.5 - 1) times 1.61.5 - 1.6$$\frac{f_m}{20} = \frac{(1.5 - 1) \times 1.6}{1.5 - 1.6}$$
### Step 1: Simplify and Compute
$fracf_m20 = frac0.5 times 1.6-0.1$$$\frac{f_m}{20} = \frac{0.5 \times 1.6}{-0.1}$$fracf_m20 = frac0.8-0.1 = -8$$\frac{f_m}{20} = \frac{0.8}{-0.1} = -8$$f_m = -8 times 20 = -160 mathrm~cm$$f_m = -8 \times 20 = -160 \mathrm{~cm}$$
Therefore, the focal length in the liquid is -160 mathrm~cm$-160 \mathrm{~cm}$.
### Pattern Recognition
Notice that since the surrounding liquid medium has a higher refractive index than the lens material itself (mu_m gt mu_1$\mu_m \gt \mu_1$), the sign of the focal length flips from positive to negative. The convex lens behaves as a diverging lens inside this specific liquid.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q54jee_main_2024_30_january_eveningLens Formula and Displacement
In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The y-x plot is shown in figure.
The focal length of the lens is ________ mathrmcm$\mathrm{cm}$.
A graph showing y versus x with a curve passing through the point (20, 20).
Numerical Answer.Answer: 20 to 20
Solution
### Related Formula
textNewton's Lens Formula: x_1 x_2 = f^2$$\text{Newton's Lens Formula: } x_1 x_2 = f^2$$
where x_1$x_1$ and x_2$x_2$ are object and image distances from the focal points.
### Core Logic
Since distances x$x$ and y$y$ are measured from the focal point (not the optical center), we use Newton's formula: x y = f^2$x y = f^2$.
From the graph, a prominent point on the curve is (20, 20)$(20, 20)$.
### Step 1: Calculate Focal Length
20 times 20 = f^2$$20 \times 20 = f^2$$f^2 = 400 implies f = 20 mathrm~cm$$f^2 = 400 \implies f = 20 \mathrm{~cm}$$
Alternatively using standard lens formula:
Object distance from optical center u = -(f + x)$u = -(f + x)$
Image distance v = +(f + y)$v = +(f + y)$frac1v - frac1u = frac1f$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$frac1f+y - frac1-(f+x) = frac1f$$\frac{1}{f+y} - \frac{1}{-(f+x)} = \frac{1}{f}$$
If x = y = 20 mathrm~cm$x = y = 20 \mathrm{~cm}$:
frac1f+20 + frac1f+20 = frac1f$$\frac{1}{f+20} + \frac{1}{f+20} = \frac{1}{f}$$frac2f+20 = frac1f$$\frac{2}{f+20} = \frac{1}{f}$$2f = f + 20 implies f = 20 mathrm~cm$$2f = f + 20 \implies f = 20 \mathrm{~cm}$$
### Pattern Recognition
Whenever distances are specified relative to the focal point, Newton's formula (xy = f^2$xy = f^2$) instantly solves the problem.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q55jee_main_2024_30_jan_morningLens Formula and Magnification
The distance between object and its two times magnified real image as produced by a convex lens is 45 mathrm~cm$45 \mathrm{~cm}$. The focal length of the lens used is \_ \_ \_ \_ \_ \_ mathrmcm$\_ \_ \_ \_ \_ \_ \mathrm{cm}$
Numerical Answer.Answer: 10 to 10
Solution
### Related Formula
m = fracvu$$m = \frac{v}{u}$$frac1f = frac1v - frac1u$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$
### Core Logic
For a real image produced by a convex lens, the magnification m$m$ is negative. The distance between the object and the real image is the absolute sum of their distances from the lens: |v| + |u| = v - u = 45 mathrm~cm$|v| + |u| = v - u = 45 \mathrm{~cm}$ (since u$u$ is negative and v$v$ is positive).
### Step 1: Set Up Magnification and Distance
fracvu = -2$$\frac{v}{u} = -2$$v = -2u quad dots (i)$$v = -2u \quad \dots (i)$$
Distance between object and image:
v - u = 45 quad dots (ii)$$v - u = 45 \quad \dots (ii)$$
### Step 2: Solve for Object and Image Distances
Substitute (i) into (ii):
(-2u) - u = 45$(-2u) - u = 45$-3u = 45 Rightarrow u = -15 mathrm~cm$$-3u = 45 \Rightarrow u = -15 \mathrm{~cm}$$
Then, v = -2(-15) = +30 mathrm~cm$v = -2(-15) = +30 \mathrm{~cm}$.
### Step 3: Calculate Focal Length
frac1f = frac1v - frac1u$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$frac1f = frac130 - frac1-15$$\frac{1}{f} = \frac{1}{30} - \frac{1}{-15}$$frac1f = frac130 + frac230 = frac330 = frac110$$\frac{1}{f} = \frac{1}{30} + \frac{2}{30} = \frac{3}{30} = \frac{1}{10}$$f = +10 mathrm~cm$$f = +10 \mathrm{~cm}$$
### Pattern Recognition
For a real image, total object-to-image distance D = v + |u|$D = v + |u|$. Applying sign convention naturally resolves this to D = v - u$D = v - u$. Two conditions (magnification + total distance) reliably solve for both u, v$u, v$ before hitting the lens equation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q52jee_main_2024_31_jan_eveningRefraction at Spherical Surfaces
Light from a point source in air falls on a convex curved surface of radius 20 text cm$20 \text{ cm}$ and refractive index 1.5. If the source is located at 100 text cm$100 \text{ cm}$ from the convex surface, the image will be formed at ______ cm from the object.
Numerical Answer.Answer: 200 to 200
Solution
### Related Formula
fracmu_2v - fracmu_1u = fracmu_2 - mu_1R$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$
### Core Logic
For a convex refracting surface, the radius of curvature R$R$ is positive if the center of curvature lies in the denser medium.
mu_1 = 1 text (air)
$
\mu_1 = 1 \text{ (air)}
$
mu_2 = 1.5 text (glass/medium)
$
\mu_2 = 1.5 \text{ (glass/medium)}
$
u = -100 text cm
$
u = -100 \text{ cm}
$
R = +20 text cm
$
R = +20 \text{ cm}
$Refraction at Spherical Surfaces diagram for Q52 - JEE Main 2024 Evening
### Step 1: Calculate Image Position
frac1.5v - frac1-100 = frac1.5 - 120$$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20}$$frac1.5v + frac1100 = frac0.520$$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20}$$frac1.5v = frac140 - frac1100$$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100}$$frac1.5v = frac5 - 2200 = frac3200$$\frac{1.5}{v} = \frac{5 - 2}{200} = \frac{3}{200}$$v = frac1.5 times 2003 = 100 text cm$$v = \frac{1.5 \times 200}{3} = 100 \text{ cm}$$
### Step 2: Distance from Object
The image is formed at v = +100 text cm$v = +100 \text{ cm}$ from the pole, which is on the other side of the surface.
The object is at |u| = 100 text cm$|u| = 100 \text{ cm}$ from the pole.
Distance between object and image = |u| + v = 100 + 100 = 200 text cm$= |u| + v = 100 + 100 = 200 \text{ cm}$.
### Pattern Recognition
Always read the final line of optics questions carefully. The question asks for distance "from the object" not "from the surface/pole". This is a classic trap where students answer 100 instead of 200.
### 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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