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 Type:
Enter a numerical valueAnswer: 100 to 100+4 marks
Solution & Explanation
### 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
Keywords:#In a microscope the objective is having#JEE Main 2026 Morning Q50#Ray Optics and Optical Instruments JEE Main 2026#Microscope JEE Main 2026
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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