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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will

(a) Act as a convex lens only for the objects that lie on its curved side

(b) Act as a concave lens for the objects that lie on its curved side

(c) Act as a convex lens irrespective of the side on which the object lies

(d) Act as a concave lens irrespective of side on which the object lies

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

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Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

You are given four sources of light each one providing a light of a single colour—red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?

(a) The beam of red light would undergo total internal reflection.

(b) The beam of red light would bend towards the normal while it gets refracted through the second medium.

(c) The beam of blue light would undergo total internal

...more

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) According to VIBGYOR, among all given sources of light, the blue light have the smallest wavelength. According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle. So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A passenger in an aeroplane shall

(a) Never see a rainbow

(b) May see a primary and a secondary rainbow as concentric circles

(c) May see a primary and a secondary rainbow as concentric arcs

(d) Shall never see a secondary rainbow

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) A passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.

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Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image

(a) Moves away from the lens with an uniform speed 5 m/s

(b) Moves away from the lens with an uniform acceleration

(c) Moves away from the lens with a non-uniform acceleration

(d) Moves towards the lens with a non-uniform acceleration

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) When an object approaches a convergent lens from the left of the lens with a uniform speed of 5 m/s, the image away from the lens with a non-uniform acceleration.

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A' short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is

(a) Blue

(b) Green

(c) Violet

(d) Red

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(d) Since v ∝ λ, the light of red colour is of the highest wavelength and therefore of the highest speed. Therefore, after travelling through the slab, the red colour emerge first.

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A ray of light incident at an angle d on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is mad? of a material of refractive index 1.5, the angle of incidence is

(a) 7.5°

(b) 5°

(c) 15°

(d) 2.5°

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) since deviation $δ = (μ - 1) A$ = (1.5-1)5⁰= 2.5⁰

But $δ = θ - r$ so $θ = 7.5 °$

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

In many experimental set-ups, the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

1/f=1/v-1/u

-u+V=D

U=-(D-v)

Putting $\frac{1}{D - v} + \frac{1}{v} = \frac{1}{f}$

On solving v²-Dv+Dt=0

So v= $\frac{D}{2} ? \frac{\sqrt[]{D^{2} - 4 D f}}{2}$

When the objects distance is $\frac{D}{2} + \frac{\sqrt[]{D^{2} - 4 D t}}{2}$

The image forms at $\frac{D}{2} ? \frac{\sqrt[]{D^{2} - 4 D f}}{2}$

Similarly when the objects distance is $\frac{D}{2} + \frac{\sqrt[]{D^{2} - 4 D t}}{2}$

The image forms at $\frac{D}{2} + \frac{\sqrt[]{D^{2} - 4 D t}}{2}$

The distance between the poles for these two objects distance is

\frac{D}{2} + \frac{\sqrt[]{D^{2} - 4 D t}}{2}

\frac{D}{2} - \frac{\sqrt[]{D^{2} - 4 D t}}{2}

\sqrt[]{D^{2} - 4 D t}

Let d =

\sqrt[]{D^{2} - 4 D t}

If u = D/2+d/2 then the image is at v=D/2-d/2

The magnification m₁= $\frac{D + d}{D - d}$

If u =D-d/2then v= D+d/2

The magnification m₂= $\frac{D + d}{D - d}$

$\frac{m_{2}}{m_{1}} = \frac{{(D + d)}^{2}}{{(D - d)}^{2}}$

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0, 0) and an object is placed at (-50 cm, 0). Find the coordinates of the image.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

If there was no cut, then the object would have been at a height of 0.5 cm from the principal axis OO'.

Applying lens formula, we have

1/v-1/u=1/f $\frac{1}{D - v} + \frac{1}{v} = \frac{1}{f}$ Z

= $\frac{1}{- 50} + \frac{1}{25}$

V= 50cm

Magnification m = v/u= 50/-50=-1

So coordinates of image are (50cm, -1cm)

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- p=100W

_{$λ$
1}= 1nm

_{$λ$
2}= 500nm

Also n₁and n₂ represent number of photon of x-rays and visible light emitted from the two sources per sec

E/t=P=n₁hc/ $λ$ ₁₌n₂hc/ $λ$ ₂

So n₁/ $λ$ ₁= n₂/ $λ$ ₂

So n₁/n₂= 1/500

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A circular disc of radius R is placed co-axially and horizontally inside an opaque hemispherical bowl of radius a (figure). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index μ. and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Refering to the figure, AM is the direction of incidence ray before liquid is filled. After liquid is filled in, BM is the direction of the incident ray. Refracted ray in both cases is same as that along AM

N= 90⁰, OM= a, CB = NB= a-R, AN= a+R

Sint=

\frac{a - R}{\sqrt[]{d^{2} + ({a - R)}^{2})}}

Sin

\propto

= cos 90-

\propto

\frac{a + R}{\sqrt[]{d^{2} + ({a + R)}^{2})}}

By applying snells law

\frac{1}{\propto} = \frac{s i n t}{s i n r} = \frac{s i n t}{s i n \propto}

On substituting the values d =

\frac{\propto (a^{2} - b^{2})}{\sqrt[]{{(a + r)}^{2} - \propto ({a - r)}^{2}}}

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