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Physics NCERT Exemplar Solutions Class 12th Chapter Nine

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

Three immiscible liquids of densities d₁> d₂> d₃and refractive indices μ₁> μ₂> μ₃ are put in a beaker. The height of each liquid column is h/3. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

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Payal Gupta

Contributor-Level 10

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

apparent depth = real depth / refractive index

Since, the image formed by Medium1, O₂act as an object for Medium2.

If seen from µ₃ ,the apparent depth is O₂.

Similarly, the image formed by Medium2, O₂ act as an object for Medium3

O₂₌ $\frac{μ_{3}}{μ_{2}}$ ( $\frac{h}{3} + O_{1}$ )

= $\frac{μ_{3}}{μ_{2}}$ ( $\frac{h}{3} + \frac{μ_{2}}{μ_{1}} \frac{h}{3})$

Seen from outside the apparent height is

O₃ = $\frac{1}{μ_{3}}$ ( $\frac{h}{3} + O_{2})$ = $\frac{1}{μ_{3}} (\frac{h}{3} + \frac{h}{3} (\frac{μ_{3}}{μ_{2}} + \frac{μ_{3}}{μ_{1}}))$

= $\frac{h}{3} (\frac{1}{μ_{1}} + \frac{1}{μ_{2}} + \frac{1}{μ_{3}}$ )

This is the expression for required depth.

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

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Payal Gupta

Contributor-Level 10

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

no reversibility of lens maker formula is not possible.

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

The near vision of an average person is 25 cm. To view an object with an angular magnification of 10, what should be the power of the microscope?

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Payal Gupta

Contributor-Level 10

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

m=v/u = D/f

So m = 25/10 = 2.5= 0.025m

P= 1/0.025= 40D

This is the required power of lens.

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

Will the focal length of a lens for red light be more, same or less than that for blue light?

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Payal Gupta

Contributor-Level 10

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

As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.

By lens makers formula 1/f= $(μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

So f_br so focal length of blue is less than red light

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

(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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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

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

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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.

less

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

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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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

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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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

Show that for a material with refractive index μ ≥ √2, light incident at angle shall be guided along, a length perpendicular to the incident face.

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Payal Gupta

Contributor-Level 10

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

Any ray entering at an angle I shall be guided along AC if the angle ray makes with the face AC (φ) is greater than the critical angle as per the principle of total internal reflection φ +r =900, therefore sinφ = cosr

Sinφ> $\frac{1}{μ}$

Cosr> $\frac{1}{μ}$

1-cos²r<1-1/ $μ^{2}$

Sin²r<1-1/ $μ^{2}$

Sini = $μ$ sinr

I= $\frac{π}{2}$

If that is greater than the critical then all other angle of incidence shall be more than the critical angle.

1< $μ^{2}$ -1

$μ$ > $2$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Class 12th

A myopic adult has a far point at 0.1 m. His power of accomodation is 4 diopters.

(i) What power lenses are required to see distant objects?

(ii) What is his near point without glasses?

(iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)

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Payal Gupta

Contributor-Level 10

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

(i) As we know that power P_f=1/f=1/0.1+1/0.02=60D

By corrective lens the object distance at far pointP'_f=1/f'=1 $\infty +$ /1/0.02=50D

Total power P'_f=P_f+P_g

P_g=-10D

(ii) This power of accommodation is 4D for the normal eye then

4=P_n-P_f where P_n power of near point

So P_n= 64D

1/x_n+1/0.002=64 then x_n= 1/14=0.07m

(iii) P_n'=P_f'+4=54

After solving x_n'=4=0.25m

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