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 2
Q4jee_main_2025_02_april_morningRefraction through Lenses
A slanted object AB is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is:
A slanted object AB forming an angle alpha with the principal axis of a convex lens.
A.-fracalpha2$-\frac{\alpha}{2}$
B.-45^circ$-45^{\circ}$
C.+45^circ$+45^{\circ}$
D.-alpha$-\alpha$
Solution
### Related Formula
frac1v - frac1u = frac1f$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$m = fracvu$$m = \frac{v}{u}$$m_L = fracdvdu = m^2$$m_L = \frac{dv}{du} = m^2$$
### Core Logic
Let the pole of the lens be the origin. Point A$A$ of the slanted object lies on the principal axis at u = -30mathrm~cm$u = -30\mathrm{~cm}$ from the convex lens (f = +20mathrm~cm$f = +20\mathrm{~cm}$).
Let's locate the image of A$A$:
frac1v - frac1-30 = frac120 implies frac1v = frac120 - frac130 = frac160 implies v = +60mathrm~cm$$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \implies v = +60\mathrm{~cm}$$
Thus, the transverse magnification m$m$ at point A$A$ is:
m = fracvu = frac60-30 = -2$$m = \frac{v}{u} = \frac{60}{-30} = -2$$
Since the longitudinal extension of the object is small (du = 1mathrm~cm$du = 1\mathrm{~cm}$ along the axis):
dv = m^2 du = (-2)^2 times 1 = 4mathrm~cm$$dv = m^2 du = (-2)^2 \times 1 = 4\mathrm{~cm}$$
The height of the object at point B$B$ is h_o = 2mathrm~cm$h_o = 2\mathrm{~cm}$. Its image height is:
h_i = m cdot h_o = (-2) times 2 = -4mathrm~cm$$h_i = m \cdot h_o = (-2) \times 2 = -4\mathrm{~cm}$$
Now, compute the angle beta$\beta$ made by the image with the principal axis:
tanbeta = frach_idv = frac-4mathrm~cm4mathrm~cm = -1$$\tan\beta = \frac{h_i}{dv} = \frac{-4\mathrm{~cm}}{4\mathrm{~cm}} = -1$$beta = -45^circ$$\beta = -45^{\circ}$$
### Step 1: Final Conclusion
The angle made by the image with the principal axis is $
### Step 1: Final Conclusion
The angle made by the image with the principal axis is $-45^{\circ}.
### Pattern Recognition
For small objects tilted with respect to the principal axis:
1. Axial displacement scales by $.
### Pattern Recognition
For small objects tilted with respect to the principal axis:
1. Axial displacement scales by $m^2.
2. Transverse height scales by $.
2. Transverse height scales by $m.
3. Slope scales by $.
3. Slope scales by $\frac{m}{m^2} = \frac{1}{m}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics
Q17jee_main_2025_02_april_morningRefraction at Spherical Surfaces
A spherical surface separates two media of refractive indices 1 and 1.5 as shown in the figure. Distance of the image of an object 'O', is:
(C is the center of curvature of the spherical surface and R is the radius of curvature)
A spherical surface of radius 0.4 m separating media of n1 = 1 and n2 = 1.5, with object O at 0.2 m.
A.0.24mathrm~m$0.24\mathrm{~m}$ right to the spherical surface
B.0.4mathrm~m$0.4\mathrm{~m}$ left to the spherical surface
C.0.24mathrm~m$0.24\mathrm{~m}$ left to the spherical surface
D.0.4mathrm~m$0.4\mathrm{~m}$ right to the spherical surface
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
From the given diagram, using the standard Cartesian sign convention with the pole of the surface as origin:
- Refractive index of first medium, $
### Core Logic
From the given diagram, using the standard Cartesian sign convention with the pole of the surface as origin:
- Refractive index of first medium, $\mu_1 = 1.0
- Refractive index of second medium, $
- Refractive index of second medium, $\mu_2 = 1.5
- Object distance, $
- Object distance, $u = -0.2\mathrm{~m} (left of surface)
- Radius of curvature, $ (left of surface)
- Radius of curvature, $R = +0.4\mathrm{~m} (convex surface towards first medium, center $ (convex surface towards first medium, center $C lies in second medium)
Applying the formula for refraction at a spherical interface:
$ lies in second medium)
Applying the formula for refraction at a spherical interface:
$frac1.5v - frac1-0.2 = frac1.5 - 10.4$\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4}$$
$frac1.5v + 5.0 = frac0.50.4 = 1.25$\frac{1.5}{v} + 5.0 = \frac{0.5}{0.4} = 1.25$$
$frac1.5v = 1.25 - 5.0 = -3.75$\frac{1.5}{v} = 1.25 - 5.0 = -3.75$$
$v = frac1.5-3.75 = -0.4mathrm~m$v = \frac{1.5}{-3.75} = -0.4\mathrm{~m}$
The negative sign indicates that the image is formed to the left of the spherical refracting surface.
### Step 1: Final Conclusion
The image of object 'O' is formed $
The negative sign indicates that the image is formed to the left of the spherical refracting surface.
### Step 1: Final Conclusion
The image of object 'O' is formed $0.4\mathrm{~m}$ left to the spherical surface.
### Pattern Recognition
Ensure to strictly implement the coordinate sign conventions: the direction of incident light is positive. Since light goes from left to right, left-side points have a negative coordinate, and right-side points have a positive coordinate.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics
Qjee_main_2025_03_april_eveningRefraction at Spherical Surfaces
Light from a point source in air falls on a spherical glass surface (refractive index, mu=1.5$\mu=1.5$ and radius of curvature =50$=50$ cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ________ m.
Numerical Answer.Answer: 4 to 4
Solution
### Related Formula
The refraction equation at a single spherical interface is given by:
fracmu_2v - fracmu_1u = fracmu_2 - mu_1R$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$
where:
- mu_1$\mu_1$ is the refractive index of the initial medium (air, mu_1 = 1.0$\mu_1 = 1.0$)
- mu_2$\mu_2$ is the refractive index of the second medium (glass, mu_2 = 1.5$\mu_2 = 1.5$)
- u$u$ is the object distance
- v$v$ is the image distance
- R$R$ is the radius of curvature
### Core Logic
Given parameters:
- mu_1 = 1.0$\mu_1 = 1.0$, mu_2 = 1.5$\mu_2 = 1.5$
- Radius of curvature R = +50mathrm~cm$R = +50\mathrm{~cm}$
- Image distance v = +200mathrm~cm$v = +200\mathrm{~cm}$ (real image inside glass)
Refraction at Spherical Surfaces
### Step 1: Substitute parameters into refraction formula
$frac1.5200 - frac1u = frac1.5 - 1.050$$$\frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1.0}{50}$$frac3400 - frac1u = frac0.550 = frac1100$$\frac{3}{400} - \frac{1}{u} = \frac{0.5}{50} = \frac{1}{100}$$
### Step 2: Solve for object distance ($
### Step 2: Solve for object distance ($u)
$)
$-frac1u = frac1100 - frac3400$-\frac{1}{u} = \frac{1}{100} - \frac{3}{400}$$
$-frac1u = frac4 - 3400 = frac1400$-\frac{1}{u} = \frac{4 - 3}{400} = \frac{1}{400}$$
$u = -400mathrm~cm = -4mathrm~m$u = -400\mathrm{~cm} = -4\mathrm{~m}$
The magnitude of the distance of the light source is $
The magnitude of the distance of the light source is $4\mathrm{~m}.
### Pattern Recognition
Remember standard sign convention: Light travels from object to refracting surface. Distances measured in the direction of incident light are positive. Here, $.
### Pattern Recognition
Remember standard sign convention: Light travels from object to refracting surface. Distances measured in the direction of incident light are positive. Here, $v and $ and $R are positive, whereas $ are positive, whereas $u$ is negative.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q9jee_main_2025_03_april_eveningRefraction of Light and Refractive Index
A monochromatic light of frequency 5times10^14mathrm~Hz$5\times10^{14}\mathrm{~Hz}$ travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be :
A. 300 nm
B. 600 nm
C. 400 nm
D. 500 nm
Solution
### Related Formula
For light propagation, wave velocity, frequency, and wavelength are related by:
v = f lambda Rightarrow lambda_textair = fraccf$$v = f \lambda \Rightarrow \lambda_{\text{air}} = \frac{c}{f}$$
When light passes into a medium of refractive index mu$\mu$, the frequency remains constant, but the wavelength scales down to:
lambda_textmedium = fraclambda_textairmu$$\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu}$$
### Core Logic
Given parameters:
- Frequency f = 5 times 10^14mathrm~Hz$f = 5 \times 10^{14}\mathrm{~Hz}$
- Speed of light in vacuum/air c approx 3 times 10^8mathrm~m/s$c \approx 3 \times 10^8\mathrm{~m/s}$
- Refractive index of medium mu = 2$\mu = 2$
### Step 1: Calculate Wavelength in Air (Vacuum)
lambda_textair = frac3 times 10^8mathrm~m/s5 times 10^14mathrm~Hz = 0.6 times 10^-6mathrm~m = 600mathrm~nm$$\lambda_{\text{air}} = \frac{3 \times 10^8\mathrm{~m/s}}{5 \times 10^{14}\mathrm{~Hz}} = 0.6 \times 10^{-6}\mathrm{~m} = 600\mathrm{~nm}$$
### Step 2: Calculate Refracted Wavelength in Medium
lambda_textmedium = fraclambda_textairmu = frac600mathrm~nm2 = 300mathrm~nm$$\lambda_{\text{medium}} = \frac{\lambda_{\text{air}}}{\mu} = \frac{600\mathrm{~nm}}{2} = 300\mathrm{~nm}$$
### Pattern Recognition
Remember: Frequency is a source characteristic and never changes during refraction. Speed and wavelength both decrease by a factor of mu$\mu$ inside the medium.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Qjee_main_2025_07_april_morningRefraction at Plane Surfaces
A container contains a liquid with refractive index of 1.2 up to a height of 60~mathrmcm$60~\mathrm{cm}$ and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40~mathrmcm$40~\mathrm{cm}$ . The value of H is ______mathrmcm$\mathrm{cm}$ . (Consider liquids are immisible)
Numerical Answer.Answer: 80 to 80
Solution
### Related Formula
The apparent shift Delta x$\Delta x$ in depth through multiple immiscible liquid layers viewed normally is the sum of the individual layer shifts:
Delta x = sum d_i left( 1 - frac1mu_i right)$$\Delta x = \sum d_i \left( 1 - \frac{1}{\mu_i} \right)$$Layer stack apparent depth diagram
### Core Logic
For two liquid layers:
- Layer 1: d_1 = 60 mathrm~cm$d_1 = 60 \mathrm{~cm}$, mu_1 = 1.2$\mu_1 = 1.2$
- Layer 2: d_2 = H mathrm~cm$d_2 = H \mathrm{~cm}$, mu_2 = 1.6$\mu_2 = 1.6$
Total apparent shift is given as Delta x = 40 mathrm~cm$\Delta x = 40 \mathrm{~cm}$.
### Step 1: Set Up and Solve the Equation
Substitute the parameters into the equation:
40 = 60 left( 1 - frac11.2 right) + H left( 1 - frac11.6 right)$$40 = 60 \left( 1 - \frac{1}{1.2} \right) + H \left( 1 - \frac{1}{1.6} \right)$$
Calculate the fractional factors:
1 - frac11.2 = 1 - frac56 = frac16$$1 - \frac{1}{1.2} = 1 - \frac{5}{6} = \frac{1}{6}$$1 - frac11.6 = 1 - frac58 = frac38$$1 - \frac{1}{1.6} = 1 - \frac{5}{8} = \frac{3}{8}$$
Substitute back:
40 = 60 left( frac16 right) + H left( frac38 right)$$40 = 60 \left( \frac{1}{6} \right) + H \left( \frac{3}{8} \right)$$40 = 10 + frac38 H implies frac38 H = 30$$40 = 10 + \frac{3}{8} H \implies \frac{3}{8} H = 30$$H = frac30 times 83 = 80 mathrm~cm$$H = \frac{30 \times 8}{3} = 80 \mathrm{~cm}$$
### Pattern Recognition
Sees: Two immiscible liquid layers with normal viewing shift.
Shortcut: First layer has real depth 60, index 1.2 $
### Pattern Recognition
Sees: Two immiscible liquid layers with normal viewing shift.
Shortcut: First layer has real depth 60, index 1.2 $\implies apparent shift is $ apparent shift is $60 \times (1 - 5/6) = 10 \mathrm{~cm}. Since total shift is 40, the second layer must contribute $. Since total shift is 40, the second layer must contribute $30 \mathrm{~cm} of shift. Thus, $ of shift. Thus, $H \times (1 - 5/8) = 30 \implies H \times (3/8) = 30 \implies H = 80 \mathrm{~cm}$.
### 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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