Geometrical Optics — Page 10 | JEE Physics

Q91

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of

f_1

in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of

f_2

when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5 , the ratio of

f_1

and

f_2

will be

A

2: 3

B

1: 3

C

1: 2

D

3: 5

Correct Answer

Option B

Solution

\frac{1}{f} = \left(\frac{n}{n'} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)

For a thin plano-convex lens, one surface is flat, meaning

R_2 = \infty

and

\frac{1}{R_2} = 0.

The formula reduces to:

\frac{1}{f} = \left(\frac{n}{n'} - 1\right)\frac{1}{R_1}

Thus, the focal length is:

f = \frac{R_1}{\left(\frac{n}{n'} - 1\right)}

For the First Lens (in Air) Refractive index of lens,

n = 1.5

Surrounding medium (air),

n' = 1

Radius of curvature,

R_1 = 2 \text{ cm}

Substitute into the formula:

f_1 = \frac{2}{\left(\frac{1.5}{1} - 1\right)} = \frac{2}{(1.5 - 1)} = \frac{2}{0.5} = 4 \text{ cm}

For the Second Lens (in Liquid) Refractive index of lens,

n = 1.5

Surrounding medium (liquid),

n' = 1.2

Radius of curvature,

R_1 = 3 \text{ cm}

Substitute into the formula:

f_2 = \frac{3}{\left(\frac{1.5}{1.2} - 1\right)}

First, calculate the term inside the parentheses:

\frac{1.5}{1.2} = 1.25 \quad \Rightarrow \quad 1.25 - 1 = 0.25

Now, compute

f_2

:

f_2 = \frac{3}{0.25} = 12 \text{ cm}

Ratio of the Focal Lengths

\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}

Thus, the ratio of

f_1

to

f_2

is:

1 : 3

Q92

A photograph of a landscape is captured by a drone camera at a height of 18 km . The size of the camera film is

2 \mathrm{~cm} \times 2 \mathrm{~cm}

and the area of the landscape photographed is

400 \mathrm{~km}^2

. The focal length of the lens in the drone camera is :

A 2.8 cm

B 0.9 cm

C 2.5 cm

D 1.8 cm

Correct Answer

Option D

Solution

\begin{aligned} &\mathrm{H}=18 \mathrm{~km}\\ &\text{ Size of camera film }=2 \mathrm{~cm} \times 2 \mathrm{~cm}\\ &\begin{aligned} & \mathrm{A}_{\text{image }}=400 \mathrm{~km}^2 \\ & \mathrm{x}=20 \times 10^3 \mathrm{~m}=2 \times 10^4 \mathrm{~m} \\ & \mathrm{y}=2 \times 10^{-2} \mathrm{~m} \\ & \frac{\mathrm{x}}{\mathrm{y}}=10^6=\frac{18 \mathrm{Km}}{\mathrm{f}} \\ & \mathrm{f}=18 \times 10^{-3} \mathrm{~m}=18 \mathrm{~mm} \\ & \mathrm{f}=1.8 \mathrm{~cm} \end{aligned} \end{aligned}

Q93

A thin prism

\mathrm{P}_1

with angle

4^{\circ}

made of glass having refractive index 1.54 , is combined with another thin prism

\mathrm{P}_2

made of glass having refractive index 1.72 to get dispersion without deviation. The angle of the prism

\mathrm{P}_2

in degrees is

A 1.5

B 16/3

C 3

D 4

Correct Answer

Option C

Solution

We know, for thin prism,

\mu = {{A + \delta } \over A}

where, $\mu$ = refractive index A = Angle of prism $\delta$ = Angle of deviation We can write,

\mu A = A + \delta

\Rightarrow \delta = (\mu - 1)A

given,

{\delta _{net}} = 0

\Rightarrow ({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2} = 0

\Rightarrow (1.54 - 1)4 - (1.72 - 1){A_2} = 0

\Rightarrow {A_2} = {{54 \times 4} \over {72}} \Rightarrow {A_2} = 3^\circ

Q94

Two thin convex lenses of focal lengths 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is:

A 5 D

B 1 D

C 20 D

D 10 D

Correct Answer

Option D

Solution

To find the power of a combination of two thin convex lenses, we use the formula for the equivalent focal length $f_{\text{eq}}$ of the lens system when the lenses are placed coaxially with a certain distance $d$ between them: $\dfrac{1}{f_{\text{eq}}} = \dfrac{1}{f_1} + \dfrac{1}{f_2} - \dfrac{d}{f_1 f_2}$ Given: $f_1 = 30 \, \text{cm}$ $f_2 = 10 \, \text{cm}$ $d = 10 \, \text{cm}$ (distance between the two lenses) First, convert the focal lengths from centimeters to meters: $f_1 = 0.3 \, \text{m}$ $f_2 = 0.1 \, \text{m}$ Plug in the values into the equation: $\dfrac{1}{f_{\text{eq}}} = \dfrac{1}{0.3} + \dfrac{1}{0.1} - \dfrac{0.1}{(0.3)(0.1)}$ Calculate each term: $\dfrac{1}{0.3} = \dfrac{10}{3} \approx 3.33$ $\dfrac{1}{0.1} = 10$ $\dfrac{0.1}{(0.3)(0.1)} = \dfrac{0.1}{0.03} = \dfrac{10}{3} \approx 3.33$ Substitute these calculated values back into the formula: $\dfrac{1}{f_{\text{eq}}} = 3.33 + 10 - 3.33$ $\dfrac{1}{f_{\text{eq}}} = 10$ Thus, the power of the lens combination is: $\text{Power} = \dfrac{1}{f_{\text{eq}}} = 10 \, \text{D}$ Therefore, the power of the lens system is 10 diopters (D).

Q95

A lens having refractive index 1.6 has focal length of 12 cm , when it is in air. Find the focal length of the lens when it is placed in water. (Take refractive index of water as 1.28)

A 355 mm

B 655 mm

C 288 mm

D 555 mm

Correct Answer

Option C

Solution

To find the focal length of a lens with a refractive index of 1.6 when placed in water, given that its focal length in air is 12 cm and the refractive index of water is 1.28, we can use the lens maker's formula: $\dfrac{1}{f} = \left(\dfrac{\mu_L}{\mu_m} - 1\right) \left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ When the lens is in air, the refractive index of the medium, $\mu_m$ , is 1.

Therefore, for the lens in air: $\dfrac{1}{12} = (1.6 - 1) \left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ Simplifying gives: $\dfrac{1}{12} = \dfrac{6}{10} \left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ Solving for $\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ : $\dfrac{1}{R_1} - \dfrac{1}{R_2} = \dfrac{10}{72}$ Now, when the lens is placed in water, $\mu_m = 1.28$ .

Substituting into the equation gives: $\dfrac{1}{f} = \left(\dfrac{1.6}{1.28} - 1\right) \left(\dfrac{10}{72}\right)$ Calculating each part: $\dfrac{1}{f} = \dfrac{32}{128} \times \dfrac{10}{72}$ $\dfrac{1}{f} = \dfrac{1}{4} \times \dfrac{10}{72}$ Hence, solving for $f$ : $f = 28.8 \, \text{cm} = 288 \, \text{mm}$ Thus, the focal length of the lens in water is 288 mm.

Q96

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A): Refractive index of glass is higher than that of air.Reason (R): Optical density of a medium is directly proportionate to its mass density which results in a proportionate refractive index.In the light of the above statements, choose the most appropriate answer from the options given below:

A Both (A) and (R) are correct and (R) is 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 but (R) is not the correct explanation of (A)

Correct Answer

Option B

Solution

Refractive index has no relation with mass density because both have different meaning.

Hence reason is incorrect.

So (A) is correct but (R) is not correct.

Q97

A mirror is used to produce an image with magnification of

\dfrac{1}{4}

. If the distance between object and its image is 40 cm, then the focal length of the mirror is ________.

A 10 cm

B 12.7 cm

C 10.7 cm

D 15 cm

Correct Answer

Option C

Solution

Magnification (m): The magnification is given by the relationship: $m = \dfrac{1}{4} = \dfrac{v}{u}$ Rearrange to find the object distance (u) in terms of the image distance (v): $u = 4v$ Distance between object and image: It is given that the sum of the object distance and the image distance is 40 cm: $v + u = 40$ Substituting $u = 4v$ into the equation gives: $v + 4v = 40$ $5v = 40$ Solving for $v$ gives: $v = 8 \, \text{cm}$ Find the object distance (u): Substitute $v = 8 \, \text{cm}$ back into $u = 4v$ : $u = 4 \times 8 = 32 \, \text{cm}$ Calculate the focal length (f): Use the mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ Substituting the values of $v$ and $u$ : $\dfrac{1}{8} + \dfrac{1}{32} = \dfrac{1}{f}$ Simplify and solve for $\dfrac{1}{f}$ : $\dfrac{1}{8} - \dfrac{1}{32} = \dfrac{1}{f}$ $\dfrac{4 - 1}{32} = \dfrac{1}{f}$ $\dfrac{3}{32} = \dfrac{1}{f}$ Therefore, the focal length is: $f = \dfrac{32}{3} = 10.7 \, \text{cm}$

Q98

Consider following statements for refraction of light through prism, when angle of deviation is minimum. A. The refracted ray inside prism becomes parallel to the base. B. Larger angle prisms provide smaller angle of minimum deviation. C. Angle of incidence and angle of emergence becomes equal. D. There are always two sets of angle of incidence for which deviation will be same except at minimum deviation setting. E. Angle of refraction becomes double of prism angle. Choose the correct answer from the options given below :

A A, B and E Only

B B, C and D Only

C B, D and E Only

D A, C and D Only

Correct Answer

Option D

Solution

When light is refracted through a prism and the angle of deviation is at its minimum, the following occurs: The refracted ray inside the prism is parallel to the base of the prism.

The angle of incidence and the angle of emergence are equal.

There are two sets of angles of incidence that result in the same angle of deviation, except when the deviation is at its minimum.

These points are illustrated by the equation: $\delta = \mathrm{I} + \mathrm{e} - \mathrm{A}$ For the minimum angle of deviation ( $\delta_{\text{min}}$ ), the condition $\mathrm{I} = \mathrm{e}$ is met, and the refracted ray becomes parallel to the base.

Thus, statements A, C, and D are correct.

Q99

The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm . The refractive index of the lens material is

A 1.4

B 1.8

C 1.5

D 1.2

Correct Answer

Option C

Solution

To determine the refractive index ( $\mu$ ) of a thin convex lens, we use the lens maker's formula: $\dfrac{1}{f} = (\mu - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$ Given: Focal length ( $f$ ) = 12 cm Radii of curvature ( $R_1$ and $R_2$ ) = 10 cm and -15 cm, respectively Substituting these values into the formula, we have: $\dfrac{1}{12} = (\mu - 1) \left( \dfrac{1}{10} - \dfrac{1}{-15} \right)$ Simplifying the equation: $\dfrac{1}{12} = (\mu - 1) \left( \dfrac{3 + 2}{30} \right)$ $\dfrac{1}{12} = (\mu - 1) \times \dfrac{5}{30}$ $\dfrac{1}{12} = (\mu - 1) \times \dfrac{1}{6}$ Solving for $\mu$ : $\mu - 1 = \dfrac{1}{12} \times 6$ $\mu - 1 = \dfrac{1}{2}$ $\mu = \dfrac{3}{2}$ Thus, the refractive index of the lens material is $\mu = 1.5$ .

Q100

A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm . A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is :

A 45 cm

B 15 cm

C 22.5 cm

D 7.5 cm

Correct Answer

Option D

Solution

\begin{aligned} &\text{ For Convex mirror }\\ &\begin{aligned} & \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{30}=\frac{1}{30} \\ & \frac{1}{\mathrm{v}}=\frac{2}{30}=\frac{1}{15} \Rightarrow \mathrm{v}=15 \mathrm{~cm} \end{aligned} \end{aligned}

Image formed by convex mirror is at 45 cm from object so plane mirror should be placed midway at 22.5 cm from object so that both of their images may coinside, Therefore distance between both mirrors

=30-22.5=7.5 \mathrm{~cm}

Correct Answer : Option 2