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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Two particles A and B of de-Broglie wavelengths A, and combine to form a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one-dimensional)

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alok kumar singh

Contributor-Level 10

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

Explanation- from conservation of momentum

P_c=P_A+P_B

= $\frac{h}{λ_{c}} = \frac{h}{λ_{A}} + \frac{h}{λ_{B}}$

$\frac{h}{λ_{c}} = \frac{h λ_{A} + h λ_{B}}{λ_{A} λ_{B}}$

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case 1 when both momentum are positive

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case 2 when both momentum are negative

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case 3 when 1 is negative and 2 is positive

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{B} - λ_{A}}$

Case4 when 1 is positive and 2 is negative

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{A} - λ_{B}}$

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

(i) Consider a thin lens placed between a source (S) and an observer (O) (Figure). Let the thickness of the lens vary as w(b) =w₀– b²/α , where b is the vertical distance from the pole, w₀is a constant. Using Fermat's principle, i.e., the time of transit fora ray between the source and observer is an extremum find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.

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Payal Gupta

Contributor-Level 10

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

Time from S to P₁ is

t₁= $\frac{S P_{1}}{c} = \frac{\sqrt[]{u^{2} + b^{2}}}{c}$

$\frac{u}{c} (1 + \frac{1}{2} \frac{b^{2}}{u^{2}})$

Time from P₁ to O is

t₂= $\frac{P_{1} O}{c} = \frac{\sqrt[]{v^{2} + b^{2}}}{c}$ ; $\frac{v}{c} (1 + \frac{1}{2} \frac{b^{2}}{v^{2}})$

the time required travel through lens is t₁= $\frac{(n - 1) w b}{c}$

so total time is t=1/c(u+v+b²/2D+(n-1)(wo+b²/ $\propto$ ))

after solving we get $\propto = 2 (n - 1) D$

differentiating with respect to time

t=1/c(u+v+b²/2D+(n-1)K1In(K₂b))

dt/db=0=b/D-(n-1)K₁/b

b²= (n-1)K1D

b= $\sqrt[]{(n - 1) K_{1} D}$

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New answer posted

7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then, what inference can you make about their frequencies?

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alok kumar singh

Contributor-Level 10

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

Explanation- Suppose nA is the number of photons falling per second of beam A and nB is the number of photons falling per second of beam B.

N_A=2n_B

energy of falling photon A=hv_A

B=hv_B

as we know intensities are same

n_Ahv_A=n_Bhv_B

v_a/v_b=n_B/n_A=1/2

v_B=2v_A

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7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

Infinitely long cylinder of radius R is made of an unusual exotic material with refractive index -1 (figure). The cylinder is placed between two planes whose normals are along the y-direction. The centre of the cylinder O lies along they-axis. A narrow laser beam is directed along the y-direction from the lower plate. The laser source is at a horizontal distance x from the diameter in the y direction. Find the range of x such that light emitted from the lower plane does not reach the upper plane.

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Payal Gupta

Contributor-Level 10

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

Since the material is of refractive index $μ - 1$ , $θ_{r}$ is negative and $θ_{r}$ is positive

| $θ_{t}$ |= | $θ_{r}$ |=| $θ_{r}$ |

Explanation- since the material is of refractive index $μ - 1$ , $θ_{r}$ is negative and $θ_{r}$ is positive

| $θ_{t}$ |= | $θ_{r}$ |=| $θ_{r}$ |

The total deviation of the outcoming ray from the incoming ray is 4 $θ_{t}$ rays shall not receive if

$\frac{π}{2}$ <4 $θ_{t}$ < $\frac{3 π}{2}$

$o n s o l v i n g$ $\frac{π}{8}$ <4 $θ_{t}$ < $\frac{3 π}{8}$

sin $θ_{t}$ =x/R

$\frac{π}{8}$ -1x/R< $\frac{3 π}{8}$

$\frac{π}{8}$ $\frac{3 π}{8}$

Light emitted from the source shall not reach the receiving plate under this condition

...more

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

Since the material is of refractive index $μ - 1$ , $θ_{r}$ is negative and $θ_{r}$ is positive

| $θ_{t}$ |= | $θ_{r}$ |=| $θ_{r}$ |

Explanation- since the material is of refractive index $μ - 1$ , $θ_{r}$ is negative and $θ_{r}$ is positive

| $θ_{t}$ |= | $θ_{r}$ |=| $θ_{r}$ |

The total deviation of the outcoming ray from the incoming ray is 4 $θ_{t}$ rays shall not receive if

$\frac{π}{2}$ <4 $θ_{t}$ < $\frac{3 π}{2}$

$o n s o l v i n g$ $\frac{π}{8}$ <4 $θ_{t}$ < $\frac{3 π}{8}$

sin $θ_{t}$ =x/R

$\frac{π}{8}$ -1x/R< $\frac{3 π}{8}$

$\frac{π}{8}$ $\frac{3 π}{8}$

Light emitted from the source shall not reach the receiving plate under this condition.

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New answer posted

7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg uncertainty principle (∆x x∆p=h). You can assume the uncertainty in position ∆x as 1 nm. Assuming p = ∆p, find the energy of the electron in electron volts.

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alok kumar singh

Contributor-Level 10

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

Explanation-

$? p = \frac{h}{? x} = \frac{h}{2 π ? x}$

= $\frac{6.64 * 10^{- 34}}{2 * 22 / 7 * 10^{- 9}} = 1.05 * 10^{- 25}$ kgm/s

E= p²/2m= $\frac{{(1.05 * 10^{- 25})}^{2}}{2 * 9.1 * 10^{- 31}}$ J= 3.8 $* 10^{- 2}$ eV

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New answer posted

7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refrative index of the medium given by n(r)=1+2GM/rc²where r is the distance of the point of consideration from the centre of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.

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Payal Gupta

Contributor-Level 10

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

Let us consider planes r and r+dr. let the light be incident at an angle $θ$

n (r)sin $θ = n (r + d r) s i n ? (θ + d θ)$

n (r)sin $θ n (r) s i n θ + \frac{d n}{d r} d r s i n θ + n (r) + d r) c o s θ d θ$

-dn/drtan $θ = \frac{n (r) d θ}{d r}$

$\frac{2 G m}{r^{2} c^{2}} t a n θ = (1 + \frac{2 G M}{r c^{2}}) d θ / d r$

So after integral r²=x²+R²and tan $θ = \frac{R}{x}$

2rdr=2xdx

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New answer posted

7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The mixture of a pure liquid and a solution in a long vertical column (i.e., horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.

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Payal Gupta

Contributor-Level 10

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

Let us consider a portion of a ray between x and x+dx inside the liquid. Let the angle of incidence at x be θ and let it enter the thin column at height y. Because of the bending it shall emerge at x+dx with an angle θ+dθ and at a height y+dy . From Snell's law,

μ y s i n θ = μ (y + d y) s i n (θ + d θ)

μ y s i n θ (μ y + \frac{d μ}{d y} d y) s i n θ c o s d θ + c o s θ s i

θ

μ y c o s θ d θ

\frac{d μ}{d y} d y s i n θ

d $θ - \frac{d μ}{d y} d y t a n θ$

tan $θ = \frac{d x}{d y}$

$θ = - \frac{1 d μ d}{μ d y}$

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New answer posted

7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Consider a metal exposed to light the wavelength of 600 nm. The maximum energy of electron doubles when light of wavelength 400 nm is used.Find the work function in eV?

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alok kumar singh

Contributor-Level 10

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

Explanation- k'_max= 2k_max

K'_max= hc/ $λ - ?$

2K_max= hc/ $λ^{'} - ?_{0}$

2 (1230/600- $?$ )= (1230/400- $?$ )

So $?$ =1.02eV

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New answer posted

7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

0 Follower 9 Views

alok kumar singh

Contributor-Level 10

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

Explanation- During photoelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.

The excited electron is emitted. Therefore, the conservation of momentum is to be Considered as the momentum of incident photon transferred to the nucleus and electrons.

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New answer posted

7 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Consider figure for photoemission. How would you reconcile with momentum conservation ? note light have momentum in a different direction than the emitted electrons.

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alok kumar singh

Contributor-Level 10

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

Explanation- during photoelectric emission the momentum is transferred to metal. metal absorbs absorb the photon and its momentum is transferred, and excited electron emitted.

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