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 8
Qjee_main_2025_29_jan_morningRefraction
Two light beams fall on a transparent material block at point 1 and 2 with angle theta_1$\theta_{1}$ and theta_2$\theta_{2}$ , respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given: the distance between 1 and 2, d = 4sqrt3 mathrm~cm$d = 4\sqrt{3} \mathrm{~cm}$ and theta_1 = theta_2 = cos^-1left(fracn_22n_1right)$\theta_{1} = \theta_{2} = \cos^{-1}\left(\frac{n_{2}}{2n_{1}}\right)$ , where refractive index of the block n_2 >$n_{2} >$ refractive index of the outside medium n_1$n_{1}$ , then the thickness of the block is ______ cm.
The figure details dual incident light lines penetrating an index slab layer to converge perfectly at a pinpoint terminal base boundary position.
The figure details dual incident light lines penetrating an index slab layer to converge perfectly at a pinpoint terminal base boundary position.
Numerical Answer.Answer: 6 to 6
Solution
### Related Formula
n_1 sin i = n_2 sin r$$n_1 \sin i = n_2 \sin r$$
### Core Logic
The figure details dual incident light lines penetrating an index slab layer to converge perfectly at a pinpoint terminal base boundary position.
By Snell\'s law matching standard boundary normal configurations :
n_1 sin(90^circ - theta_1) = n_2 sin theta_3 implies n_1 cos theta_1 = n_2 sin theta_3$$n_1 \sin(90^{\circ} - \theta_1) = n_2 \sin \theta_3 \implies n_1 \cos \theta_1 = n_2 \sin \theta_3$$ [cite: 711, 712]
Substituting the angle identity macro given [cite: 2, 713]:
n_1 left(fracn_22n_1right) = n_2 sin theta_3 implies sin theta_3 = frac12 implies theta_3 = 30^circ$$n_1 \left(\frac{n_2}{2n_1}\right) = n_2 \sin \theta_3 \implies \sin \theta_3 = \frac{1}{2} \implies \theta_3 = 30^{\circ}$$
### Step 1: Geometrical Thickness Resolution
From the block triangles geometry :
tan 30^circ = fracd/2t implies frac1sqrt3 = fracd2t$$\tan 30^{\circ} = \frac{d/2}{t} \implies \frac{1}{\sqrt{3}} = \frac{d}{2t}$$t = fracdsqrt32 = frac4sqrt3 cdot sqrt32 = 6text cm$$t = \frac{d\sqrt{3}}{2} = \frac{4\sqrt{3} \cdot \sqrt{3}}{2} = 6\text{ cm}$$
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q53jee_main_2024_01_february_morningLenses
The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20mathrm~cm$20\mathrm{~cm}$. The focal length of the lens used is _______ mathrmcm$\mathrm{cm}$.
Numerical Answer.Answer: 15 to 15
Solution
### Related Formula
Lens magnification formula:
m = fracvu$$m = \frac{v}{u}$$
Thin lens equation:
frac1v - frac1u = frac1f$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
### Core Logic
For a virtual image formed by a convex lens, both the object and the image lie on the same side of the lens. The image distance is three times the object distance:
v = 3u$v = 3u$
The distance between the object and its virtual image is given as 20mathrm~cm$20\mathrm{~cm}$:
v - u = 20mathrm~cm implies 3u - u = 20mathrm~cm$$v - u = 20\mathrm{~cm} \implies 3u - u = 20\mathrm{~cm}$$2u = 20mathrm~cm implies u = 10mathrm~cm$$2u = 20\mathrm{~cm} \implies u = 10\mathrm{~cm}$$
Applying Cartesian sign conventions: object distance u = -10mathrm~cm$u = -10\mathrm{~cm}$ and image distance v = -30mathrm~cm$v = -30\mathrm{~cm}$.
### Step 1: Calculate Focal Length
Substitute these values into the lens formula:
frac1-30 - frac1-10 = frac1f$$\frac{1}{-30} - \frac{1}{-10} = \frac{1}{f}$$-frac130 + frac110 = frac1f$$-\frac{1}{30} + \frac{1}{10} = \frac{1}{f}$$frac-1 + 330 = frac230 = frac115 = frac1f implies f = 15mathrm~cm$$\frac{-1 + 3}{30} = \frac{2}{30} = \frac{1}{15} = \frac{1}{f} \implies f = 15\mathrm{~cm}$$
### Pattern Recognition
A virtual image from a convex lens means the object is placed inside the focal point (u < f$u < f$). This serves as a quick sanity check for your final value (10mathrm~cm < 15mathrm~cm$10\mathrm{~cm} < 15\mathrm{~cm}$). ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
If the distance between object and its two times magnified virtual image produced by a curved mirror is 15text cm$15\text{ cm}$, the focal length of the mirror must be:
A.15text cm$15\text{ cm}$
B.-12text cm$-12\text{ cm}$
C.-10text cm$-10\text{ cm}$
D.10/3text cm$10/3\text{ cm}$
Solution
### Related Formula
Magnification formula for spherical mirrors:
m = -fracvu$$m = -\frac{v}{u}$$
Mirror formula:
frac1f = frac1v + frac1u$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$
where:
* u$u$ is the object distance
* v$v$ is the image distance
* f$f$ is the focal length
### Core Logic
Since the image is magnified (m = 2$m = 2$) and virtual, the mirror must be concave (f < 0$f < 0$).
Let us use standard coordinate geometry signs: object is on the left (u$u$ is negative, say -u_0$-u_0$), virtual image is on the right (v$v$ is positive, say +v_0$+v_0$).
Given magnification:
m = 2 = -fracvu implies v = -2u$$m = 2 = -\frac{v}{u} \implies v = -2u$$
In terms of magnitudes:
v_0 = 2u_0$v_0 = 2u_0$
### Step 1: Use Distance Condition
The distance between the object and virtual image is 15text cm$15\text{ cm}$. Since the object is in front of the mirror and the virtual image is behind it:
textDistance = u_0 + v_0 = 15text cm$$\text{Distance} = u_0 + v_0 = 15\text{ cm}$$
Substitute v_0 = 2u_0$v_0 = 2u_0$:
u_0 + 2u_0 = 15 implies 3u_0 = 15 implies u_0 = 5text cm$$u_0 + 2u_0 = 15 \implies 3u_0 = 15 \implies u_0 = 5\text{ cm}$$
Thus:
* u_0 = 5text cm implies u = -5text cm$u_0 = 5\text{ cm} \implies u = -5\text{ cm}$
* v_0 = 10text cm implies v = +10text cm$v_0 = 10\text{ cm} \implies v = +10\text{ cm}$Ray diagram for virtual image in concave mirror for Q38
### Step 2: Calculate Focal Length
Using the mirror formula:
frac1f = frac1v + frac1u$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$frac1f = frac110 + frac1-5$$\frac{1}{f} = \frac{1}{10} + \frac{1}{-5}$$frac1f = frac1 - 210 = -frac110 implies f = -10text cm$$\frac{1}{f} = \frac{1 - 2}{10} = -\frac{1}{10} \implies f = -10\text{ cm}$$
Thus, the focal length is -10text cm$-10\text{ cm}$.
### Pattern Recognition
Virtual and magnified image → always concave mirror. Distance D$D$ between object and virtual image is given by D = |u| + v$D = |u| + v$. Since v = m|u|$v = m|u|$, we can simplify to |u| = fracDm+1$|u| = \frac{D}{m+1}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q33jee_main_2024_27_jan_morningPrism Deviation
If the refractive index of the material of a prism is cotleft(fracA2right)$\cot\left(\frac{A}{2}\right)$, where A$A$ is the angle of prism, then the angle of minimum deviation will be:
A.pi - 2A$\pi - 2A$
B.fracpi2 - 2A$\frac{pi}{2} - 2A$
C.pi - A$\pi - A$
D.fracpi2 - A$\frac{pi}{2} - A$
Solution
### Related Formula
mu = fracsinleft(fracA + delta_textmin2right)sinleft(fracA2right)$$\mu = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$
### Core Logic
Given mu = cotleft(fracA2right) = fraccosleft(fracA2right)sinleft(fracA2right)$\mu = \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. Equating this to the prism formula:
fraccosleft(fracA2right)sinleft(fracA2right) = fracsinleft(fracA + delta_textmin2right)sinleft(fracA2right)$$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$cosleft(fracA2right) = sinleft(fracA + delta_textmin2right)$$\cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_{\text{min}}}{2}\right)$$
### Step 1: Trigonometric Substitution
Convert the cosine term to sine:
sinleft(fracpi2 - fracA2right) = sinleft(fracA + delta_textmin2right)$$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_{\text{min}}}{2}\right)$$fracpi2 - fracA2 = fracA + delta_textmin2$$\frac{\pi}{2} - \frac{A}{2} = \frac{A + \delta_{\text{min}}}{2}$$pi - A = A + delta_textmin$$\pi - A = A + \delta_{\text{min}}$$delta_textmin = pi - 2A$$\delta_{\text{min}} = \pi - 2A$$
### Pattern Recognition
When mu$\mu$ equals a cotangent function of half-angle, the sine dynamic simplifications lead directly to linear functions involving complements of angle A$A$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Ray Optics and Optical Instruments
Q54jee_main_2024_27_jan_morningApparent Depth
Two immiscible liquids of refractive indicesfrac32$\frac{3}{2}$ and frac85$\frac{8}{5}$ respectively are put in a beaker. The height of each column is 6text cm$6\text{ cm}$. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is fracalpha4text cm$\frac{\alpha}{4}\text{ cm}$. The value of alpha$\alpha$ is ______.
Numerical Answer.Answer: 31 to 31
Solution
### Related Formula
h_textapparent = sum frach_imu_i$$h_{\text{apparent}} = \sum \frac{h_i}{\mu_i}$$
### Core Logic
Sum the contributions of both shifting mediums:
h_textapparent = frach_1mu_1 + frach_2mu_2$$h_{\text{apparent}} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$$
Given h_1 = h_2 = 6text cm$h_1 = h_2 = 6\text{ cm}$, mu_1 = frac32$\mu_1 = \frac{3}{2}$, mu_2 = frac85$\mu_2 = \frac{8}{5}$:
### Step 1: Compute fraction value
h_textapparent = frac63/2 + frac68/5 = 4 + frac308 = 4 + frac154$$h_{\text{apparent}} = \frac{6}{3/2} + \frac{6}{8/5} = 4 + \frac{30}{8} = 4 + \frac{15}{4}$$h_textapparent = frac16 + 154 = frac314text cm$$h_{\text{apparent}} = \frac{16 + 15}{4} = \frac{31}{4}\text{ cm}$$
### Step 2: Match to target format
Comparing with fracalpha4$\frac{\alpha}{4}$ directly yields:
alpha = 31$\alpha = 31$
### Pattern Recognition
Apparent depth across multi-layered planar mediums expands additively via separate individual medium thickness-to-refractive-index ratios.
### 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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