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Physics Ray Optics and Optical Instruments

What is the function of a Camera Lens?

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Jaya Sharma

Contributor-Level 10

A camera lens focuses the light onto a light-sensitive surface like film or a digital sensor for capturing an image. The camera lens gathers as much light as possible for forming a clear and bright image. This lens focuses the collected light on a specific point like camera's film or its digital sensor. The focus is achieved due to the curvature of lens and the arrangement of its optical elements. This lens refracts light rays so that they converge at focal plane where film/sensor is located.

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Physics Ray Optics and Optical Instruments

Explain the working principle of a Telescope.

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Jaya Sharma

Contributor-Level 10

A telescope functions on the working principle of gathering light form distant source and then, forming image which can be magnified for observation. Based on their design, there are two types of telescopes including refracting telescopes and reflecting telescopes. A refracting telescope has a front lens called as objective lens. This front lens is a convex lens that collects light from distant objects and then bend it to the focal point. In a reflecting telescope, a concave mirror is the primary mirror that collects light and then, reflects it to a focal point.

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Physics Ray Optics and Optical Instruments

What happens when light is refracted through a prism?

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Jaya Sharma

Contributor-Level 10

The following happens when light refracts from a prism:

When a white light passes through prism, it is separated into its constituent colours. This separation occurs because different colours of light are refracted in different amounts. This is known as dispersion.
Prism's refractive index varies with the wavelength of light. This causes shorter wavelengths to bend more than longer wavelengths. Due to the variation in bending, spread of colours is visible in rainbow.
After being dispersed within the prism, light rays reach second surface of the prism. Here, light rays again undergo refraction as they exit the prism and re-enter air.

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Physics Nuclei

How does nuclear force relate to quarks and gluons?

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Jaya Sharma

Contributor-Level 10

Nuclear force is related to quarks and gluons in the following ways:

Protons and neutrons are made up of a combination of smaller particles called quarks and down quarks. A proton has two up quarks and one down quark whereas neutron has one up quark and two down quarks.
Nuclear force is an interaction between quarks that is mediated by particles called gluons. These are the exchange particles that are important for strong interactions. In simple words, they glue quarks together as they are being exchanged between quarks constantly.
Quarks are never found alone and are often found with protons and neutrons. This happens because of the beha

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Physics Nuclei

How does the nuclear force contribute to nuclear stability?

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Jaya Sharma

Contributor-Level 10

Nuclear force is important to maintain the stability of an atomic nucleus for the following reasons:

A nucleus has positively charged sub-particles called protons. Since like charges repel each other, a nucleus, in ideal scenario, should fall apart because protons won't be able to coexist in the nucleus. However, nuclear force is extremely strong at very short range. Due to this reason, protons stay bound within nucleus and therefore, nucleus remains stable.
Nuclear force is responsible for binding energy that holds the nucleons together within a nucleus. Higher binding energy will result in a more stable nucleus.
The nuclear force shows

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Physics Nuclei

How does the nuclear force overcomes repulsion between protons?

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Jaya Sharma

Contributor-Level 10

As we all know that like charges repel each other. This means that there must be a repulsive electromagnetic force between positively charged protons. However, the nuclear force is said to be 100 times stronger than electromagnetic force which makes it strong enough to hold protons together within a nucleus. This nuclear force is much stronger than coulomb force at short ranges.

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Physics Vector Algebra

Three particles P, Q and R are moving along the vectors $\vec{A} = \hat{i} + \hat{j}, \vec{B} = \hat{j} + \hat{k}$ and $\vec{C} = - \hat{i} + \hat{j}$ respectively. They strike on point and start to move in different directions. Now particle P is moving normal to the plane which contains vector $\vec{A} a n d \vec{B} .$ Similarly particle Q is moving normal to the plane which contains vector $\vec{A} a n d \vec{C} .$ The angle between the direction of motion of P and Q $c o s^{- 1} (\frac{1}{\sqrt[]{x}}) .$ Then the value of x is_________.

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Vishal Baghel

Contributor-Level 10

Cross product will represent a vector perpendicular to both the vector.

Dir. Of Motion of P : $= (\hat{i} + \hat{j}) * (\hat{j} * \hat{k})$

= $\hat{k} - \hat{j} + \hat{i}$

Dir. Of Motion of Q = $(\hat{i} + \hat{j}) * (- \hat{i} + \hat{j})$

$= 2 \hat{k}$

$c o s θ = \frac{\vec{a} . \vec{b}}{a b} = \frac{2}{2 \sqrt[]{3}} = \frac{1}{\sqrt[]{3}} \Rightarrow θ = c o s^{- 1} (\frac{1}{\sqrt[]{3}})$

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Physics Class 12th

In a given circuit diagram, a 5 V zener diode along with a series resistance is connected across a 50 V power supply. The minimum value of the resistance required, if the maximum zener current is 90 mA be________ $Ω$ .

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Vishal Baghel

Contributor-Level 10

For 90 mA in zener diode, current in R should be greater than 90 mA.

$\frac{45}{R} = 90 * 1 0^{- 3} \Rightarrow R = \frac{45}{90 * 1 0^{- 3}} = 500 Ω$

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Physics Class 12th

In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of $5 Ω$ . However, it provides a potential difference of 1V across a load resistance of $2 Ω .$ The emf of the cell is given by $\frac{x}{10} V .$ Then the value of x is_________.

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Vishal Baghel

Contributor-Level 10

$i = \frac{ε}{R + r}$

$v_{R} = \frac{ε R}{R + r}$

$\frac{1.25 = \frac{ε (5)}{5 + r}}{1 = \frac{ε (2)}{2 + r}}$

r = 1

Using above equ : $ε = \frac{3}{2} = \frac{15}{10} V$

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Physics Mechanical Properties of Solids

The area of cross – section of a railway track is 0.01 m². The temperature variation is 10°C, Coefficient of linear expansion of material of track is 10^-⁵ / °C. The energy stored per meter in the track is_________J/m.

(Young's modulus of material of track is 10¹¹ Nm^-²)

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Vishal Baghel

Contributor-Level 10

Energy density (u) = $\frac{1}{2} * σ * ε = \frac{1}{2} * Y ε * ε = \frac{1}{2} ε^{2} Y$ $\frac{}{} \frac{}{} \frac{}{}^{}$

$\Rightarrow e n e r g y s t o r e d / m^{2} = \frac{1}{2} ε^{2} Y A$

Strain $(ε) = \frac{Δ l}{l_{0}} = α Δ T = 1 0^{- 5} * 10 = 1 0^{- 4}$ $\frac{}{_{}}^{}^{}$

$\Rightarrow e n e r g y s t o r e d / m^{2} = \frac{1}{2} * 1 0^{- 8} * 1 0^{11} * 1 0^{- 2} = 5$

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