In finding out the refractive index of the glass slab, the following observations were made through a travelling microscope: 50 vernier scale divisions (VSD), 49 MSD, 20 divisions on the main scale in each cm. For the mark on paper: \[ \text{MSR} = 8.45 \, \text{cm}, \quad \text{VC} = 26 \] For the mark on paper seen through the slab: \[ \text{MSR} = 7.12 \, \text{cm}, \quad \text{VC} = 41 \] For the powder particle on the top surface of the glass slab: \[ \text{MSR} = 4.05 \, \text{cm}, \quad \text{VC} = 1 \] (MSR = Main Scale Reading, VC = Vernier Coincidence) Refractive index of the glass slab is:
- 1.42
- 1.52
- 1.24
- 1.35
The Correct Option is A
Approach Solution - 1
1. Calculating the Least Count (LC):
\[1 \, \text{MSD} = \frac{1 \, \text{cm}}{20} = 0.05 \, \text{cm}\]
\[1 \, \text{VSD} = \frac{49}{50} \, \text{MSD} = \frac{49}{50} \times 0.05 \, \text{cm} = 0.049 \, \text{cm}\]
\[\text{LC} = 1 \, \text{MSD} - 1 \, \text{VSD} = 0.05 \, \text{cm} - 0.049 \, \text{cm} = 0.001 \, \text{cm}\]
2. For mark on paper, \( L_1 \):
\[L_1 = 8.45 \, \text{cm} + 26 \times 0.001 \, \text{cm} = 8.45 \, \text{cm} + 0.026 \, \text{cm} = 8.476 \, \text{cm}\]
3. For mark on paper seen through the slab, \( L_2 \):
\[L_2 = 7.12 \, \text{cm} + 41 \times 0.001 \, \text{cm} = 7.12 \, \text{cm} + 0.041 \, \text{cm} = 7.161 \, \text{cm}\]
4. For powder particle on the top surface, \( ZE \):
\[ZE = 4.05 \, \text{cm} + 1 \times 0.001 \, \text{cm} = 4.051 \, \text{cm}\]
5. Calculating the thickness of the slab:
\[\text{actual } L_1 = 8.476 - 4.051 = 4.425 \, \text{cm}\]
\[\text{actual } L_2 = 7.161 - 4.051 = 3.110 \, \text{cm}\]
6. Refractive index \( \mu \):
\[\mu = \frac{L_1}{L_2} = \frac{4.425}{3.110} = 1.42\]
Approach Solution -2
To find the refractive index of the glass slab, we need to establish the apparent shift that occurs when viewing through the slab. Let's break this down:
- Understanding the Observations:
- The measurements are made using a travelling microscope with certain specifications:
- \(50\) vernier scale divisions (VSD) coincide with \(49\) main scale divisions (MSD).
- Each cm of the main scale contains \(20\) divisions.
- Calculate Least Count of the Vernier:
- The formula for the least count (LC) of a vernier scale is given by:
- \(\text{LC} = \text{MSD} - \text{VSD}\)
- \(\text{MSD} = \frac{1}{20} \, \text{cm} = 0.05 \, \text{cm}\)
- \(\text{VSD} = \frac{49}{50} \times 0.05 \, \text{cm} = 0.049 \, \text{cm}\)
- Therefore, \(\text{LC} = 0.05 \, \text{cm} - 0.049 \, \text{cm} = 0.001 \, \text{cm}\)
- Determine the Actual Measurements:
- For the mark on paper:
- Total reading = \(\text{MSR} + \text{VC} \times \text{LC} = 8.45 + 26 \times 0.001 = 8.476 \, \text{cm}\)
- For the mark seen through the slab:
- Total reading = \(7.12 + 41 \times 0.001 = 7.161 \, \text{cm}\)
- For the particle on top of the glass slab (thickness):
- Total reading = \(4.05 + 1 \times 0.001 = 4.051 \, \text{cm}\)
- Calculate Real and Apparent Thickness:
- Real thickness \((t)\) of the slab = mark on paper - top particle reading = \(8.476 - 4.051 = 4.425 \, \text{cm}\)
- Apparent thickness seen through the slab = mark through slab - top particle reading = \(7.161 - 4.051 = 3.110 \, \text{cm}\)
- Calculate the Refractive Index:
- The formula for the refractive index \((\mu)\) is:
- \(\mu = \frac{\text{Real Thickness}}{\text{Apparent Thickness}} = \frac{4.425}{3.110} \approx 1.42\)
- Conclusion:
- Therefore, the refractive index of the glass slab is \(1.42\).
Thus, the correct answer is 1.42.
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Refraction of Light Questions
- Find minimum deviation in a thin prism if refractive index \(μ = cot\bigg(\frac{A}{2}\bigg)\). Here \(A\) represents the angle of prism.
- A light ray is incident on a glass slab of thickness \(4\sqrt{3}\) cm and refractive index \(\sqrt{2}\). The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of the ray after passing through the glass slab is _____ cm.
(Given \(\sin 15^\circ = 0.25\)) - For the thin convex lens, the radii of curvature are at 15 cm and 30 cm respectively. The focal length the lens is 20 cm. The refractive index of the material is :
- In a long glass tube, a mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of \(-2\), then the radius of curvature of the meniscus is:
- At the interface between two materials having refractive indices \( n_1 \) and \( n_2 \), the critical angle for reflection of an EM wave is \( \theta_c \). The \( n_1 \) material is replaced by another material having refractive index \( n_3 \), such that the critical angle at the interface between \( n_1 \) and \( n_3 \) materials is \( \theta_{c3} \). If \( n_1 > n_2 > n_3 \), \( \frac{n_2}{n_3} = \frac{2}{5} \), and \( \sin \theta_{c2} - \sin \theta_{c1} = \frac{1}{2} \), then \( \theta_{c1} \) is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.