Geometrical Optics — Page 13 | JEE Physics

Q121

A bi-convex lens has radius of curvature of both the surfaces same as

1 / 6 \mathrm{~cm}

. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides

\left(R_1 \neq R_2\right)

, without any change in lens power then possible combination of

R_1

and

R_2

is :

A

\dfrac{1}{3} \mathrm{~cm}

and

\dfrac{1}{7} \mathrm{~cm}

B

\dfrac{1}{5} \mathrm{~cm}

and

\dfrac{1}{7} \mathrm{~cm}

C

\dfrac{1}{3} \mathrm{~cm}

and

\dfrac{1}{3} \mathrm{~cm}

D

\dfrac{1}{6} \mathrm{~cm}

and

\dfrac{1}{9} \mathrm{~cm}

Correct Answer

Option B

Solution

To replace a bi-convex lens with another convex lens that has different radii of curvature on each side (i.e., $R_1 \neq R_2$ ), while maintaining the same lens power, the radii must satisfy a specific relationship.

For the current bi-convex lens, both radii of curvature are given as $\dfrac{1}{6} \, \text{cm}$ .

The lens formula for power is tied to the radii through: $\dfrac{1}{f_1} = (\mu - 1) \left( \dfrac{2}{R} \right)$ When this lens is replaced by a lens with different radii ( $R_1$ and $R_2$ ), the condition that has to be met is: $\dfrac{1}{f_2} = (\mu - 1) \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right)$ For power equivalence, the expressions for $\dfrac{1}{f_1}$ and $\dfrac{1}{f_2}$ must be equal: $(\mu - 1) \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right) = (\mu - 1) \left( \dfrac{2}{R} \right)$ This simplifies to: $\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{2}{R}$ Given $R = \dfrac{1}{6} \, \text{cm}$ , it follows that: $\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{2}{\left(\dfrac{1}{6}\right)}$ Thus, simplifying the equation gives: $\dfrac{1}{R_1} + \dfrac{1}{R_2} = 12$ Therefore, any valid pair of radii $R_1$ and $R_2$ must satisfy this equation.

Q122

Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 cm , at same distance of 18 cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is :

A 2

B

1 / 2

C

1 / 3

D 3

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Using } m=\frac{f}{u-f} \\ & m_1=\frac{6}{18-6}=\frac{1}{2} \end{aligned}

$\mathrm{m}_2=\dfrac{6}{18+6}=\dfrac{1}{4} \quad \therefore \dfrac{\mathrm{~m}_2}{\mathrm{~m}_1}=\dfrac{1}{2}$

Q123

A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid of refractive index 1.42. If its focal length in liquid is f1 , then the ratio

{{{f_1}} \over f}

is closest to the integer :

A 17

B 1

C 9

D 5

Correct Answer

Option C

Solution

Using formula

{1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

{1 \over {{f}}} = \left( {{{1.5} \over 1} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

...(1) and

{1 \over {{f_1}}} = \left( {{{1.5} \over {1.42}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

...(2) Dividing (1) by (2), we get

{{{f_1}} \over f} = {{0.5} \over {0.056}}

= 8.93 $\approx$ 9

Q124

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: The phase difference of two light waves change if they travel through different media having same thickness, but different indices of refraction. Reason R: The wavelengths of waves are different in different media. In the light of the above statements, choose the most appropriate answer from the options given below

A Both A and R are correct but R is NOT the correct explanation of A

B A is correct but R is not correct

C A is not correct but R is correct

D Both A and R are correct and R is the correct explanation of A

Correct Answer

Option D

Solution

Assertion A is correct.

When light waves travel through different media with different indices of refraction, the speed of light changes, which in turn changes the phase of the light waves.

This is because the phase of a wave is directly related to the distance it has traveled, which in this case is affected by the speed of light in the different media.

Reason R is also correct.

The speed of light in a medium is determined by its refractive index.

When light enters a medium with a different refractive index, its speed, and therefore its wavelength, changes.

This is described by the equation v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the refractive index of the medium.

Moreover, R is indeed the correct explanation of A.

The change in wavelength (and therefore phase) of the light waves in different media (as explained in R) is the reason why the phase difference changes when light waves travel through different media with the same thickness but different indices of refraction (as stated in A).

Q125

A convex lens of focal length

40 \mathrm{~cm}

forms an image of an extended source of light on a photoelectric cell. A current I is produced. The lens is replaced by another convex lens having the same diameter but focal length

20 \mathrm{~cm}

. The photoelectric current now is :

A

\mathrm{\dfrac{I}{2}}

B 4 I

C 2 I

D I

Correct Answer

Option D

Solution

As amount of energy incident on cell is same so current will remain same.

Q126

A plano convex lens of refractive index

\mu

1 and focal length ƒ1 is kept in contact with another plano concave lens of refractive index

\mu

2 and focal length ƒ2. If the radius of curvature of their spherical faces is R each and ƒ1 = 2ƒ2, then

\mu

1 and

\mu

2 are related as -

A

3{\mu _2} - 2{\mu _1}

= 1

B

{\mu _1} + {\mu _2}

= 3

C

2{\mu _1} - {\mu _2}

= 1

D

2{\mu _2} - {\mu _1}

= 1

Correct Answer

Option C

Solution

{1 \over {2{f_2}}} = {1 \over {{f_1}}} = \left( {{\mu _1} - 1} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right)

{1 \over {{f_2}}} = \left( {{\mu _2} - 1} \right)\left( {{1 \over { - R}} - {1 \over \infty }} \right)

{{\left( {{\mu _1} - 1} \right)} \over R} = {{\left( {{\mu _2} - 1} \right)} \over {2R}}

2{\mu _1} - {\mu _2} = 1

Q127

The aperture diameter of a telescope is 5m. The separation between the moon and the earth is 4 × 105 km. With light of wavelength of 5500

\mathop A\limits^o

, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to :

A 20 m

B 200 m

C 600 m

D 60 m

Correct Answer

Option D

Solution

$\theta$ = 1.22

{\lambda \over a}

Distance O1O2 = ( $\theta$ )d = (1.22

{\lambda \over a}

)d =

{{\left( {1.22} \right)\left( {5500 \times {{10}^{ - 10}}} \right)\left( {4 \times {{10}^5}} \right) \times {{10}^3}} \over 5}

= 5368 × 10–2 m = 53.68 m

Q128

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance

u

and the image distance

v,

from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of

{45^ \circ }

with the

x

-axis meets the experimental curve at

P.

The coordinates of

P

will be :

A

\left( {{f \over 2},{f \over 2}} \right)

B

\left( {f,f} \right)

C

\left( {4f,4f} \right)

D

\left( {2f,2f} \right)

Correct Answer

Option D

Solution

Here

u = - 2f,v = 2f

As

|u|

increases,

v

decreases for

|u| > f.

The graph between

|v|

and

|u|

is shown in the figure. A straight line passing through the origin and making an angle of

{45^ \circ }

with the

x

-axis meets the experimental curve at

P\left( {2f,2f} \right).

Q129

A spherical surface of radius of curvature

R

, separates air from glass (refractive index

=1.5

). The centre of curvature is in the glass medium. A point object '

O

' placed in air on the optic axis of the surface, so that its real image is formed at 'I' inside glass. The line OI intersects the spherical surface at

P

and

P O=P I

. The distance

P O

equals to

A 5R

B 2R

C 1.5R

D 3R

Correct Answer

Option A

Solution

We wish to find the distance from the point where the ray joining the object (O) and its image (I) meets the spherical surface (P) to the object.

The given condition is that

PO = PI,

meaning that P is the midpoint of OI.

Let’s work through the steps: Setup the coordinate system and geometry: • Choose the vertex of the spherical surface (the “pole”) as the origin (x = 0).

• Let the optical axis be the x‑axis.

• Since the object is in air, place O at x = –u (with u > 0).

• Since the real image is formed inside the glass, place I at x = v (with v > 0).

• The spherical surface has radius of curvature

R

and its center of curvature is in the glass. Thus, we take the center of the sphere as

C=(R,0)

.

• The vertex (the point on the sphere closest to the object) is then located at x = 0.

Position of P: • The ray joining O and I is along the x‑axis.

Its midpoint is

x_P = \frac{-u + v}{2}.

• However, since the physical part of the spherical surface is the convex cap met first by the incident light, the ray in fact meets the surface at the vertex, which is at x = 0.

• Equating the midpoint to 0, we have

\frac{-u + v}{2}=0 \quad \Longrightarrow \quad v = u.

• Thus, the object and its image are equidistant from the vertex, and

PO = |0 - (-u)| = u.

Relate u and R using the refraction at a spherical surface: • The formula for refraction at a spherical surface (using the appropriate sign–convention) is

\frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R},

where for our problem –

n_1 = 1

(air) –

n_2 = 1.5

(glass). • With

v = u

, substitute in:

\frac{1}{u} + \frac{1.5}{u} = \frac{0.5}{R} \quad \Longrightarrow \quad \frac{2.5}{u} = \frac{0.5}{R}.

• Solving for u:

u = \frac{2.5\,R}{0.5} = 5R.

Conclusion: • Since

PO = u

, we have

PO = 5R.

Thus, the distance

PO

equals

5R

, which is Option A.

Q130

A prism of refractive index

\mu

and angle of prism A is placed in the position of minimum angle of deviation. If minimum angle of deviation is also A, then in terms of refractive index

A

2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)

B

{\sin ^{ - 1}}\left( {{\mu \over 2}} \right)

C

{\sin ^{ - 1}}\left( {\sqrt {{{\mu - 1} \over 2}} } \right)

D

{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)

Correct Answer

Option A

Solution

\mu = {{\sin \left( {{{A + {\delta _{\min }}} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}

\mu = {{\sin \left( {{{A + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}

\mu = {{\sin A} \over {\sin {A \over 2}}} = 2\cos {A \over 2}

A = 2{\cos ^{ - 1}}\left( {{\mu \over 2}} \right)