Geometrical Optics Questions — NEET Physics

Q1

The magnifying power of a telescope with tube 60 cm is 5. What is the focal length of its eye piece ?

A 40 cm

B 10 cm

C 30 cm

D 20 cm

Correct Answer

Option B

Solution

For telescope Tube length (L) = f0 + fe and magnification (m) =

{{{f_0}} \over {{f_e}}}

= 5 where fo and fe are focal length of objective and eyepiece. $\therefore$ f0 + fe = 60 $\therefore$ fo = 50 cm fe = 10 cm

Q2

A ray of monochromatic light is passing through an equilateral prism

(A B C)

as shown in the figure. The refracted ray

(Q R)

is parallel to its base

(B C)

and the angle of incidence

(i)

is

50^{\circ}

. Then the angle of deviation

(\delta)

is:

A

45^{\circ}

B

35^{\circ}

C

40^{\circ}

D

55^{\circ}

Correct Answer

Option C

Solution

\begin{aligned} &i=e\\ &\text{ Equation of prism }\\ &\begin{aligned} & i+e=A+\delta \\ & 2 i-A=\delta \\ & 2(50)-60=\delta \\ & \delta=40^{\circ} \end{aligned} \end{aligned}

Q3

In a concave lens, a ray of light emanating from the object parallel to the principal axis of the lens after refraction:

A passes through

2 F

, which is the radius of curvature of the lens.

B appears to diverge from the first principal focus.

C emerges parallel to the principal axis.

D passes through the second principal focus.

Correct Answer

Option B

Solution

F_2 \text{ is the second principal focus ⇒ It is the virtual image position for object at infinity. }

$F_1$ is the first principal focus ⇒ It is the virtual object position for which image is formed at infinity.

The best appropriate answer is option (2), although it should be second principal focus.

Q4

A microscope has an objective of focal length 2 cm , eyepiece of focal length 4 cm and the tube length of 40 cm . If the distance of distinct vision of eye is 25 cm , the magnification in the microscope is

A 150

B 250

C 100

D 125

Correct Answer

Option D

Solution

To determine the magnification of the microscope, we use the formula: $m = \dfrac{L}{f_o} \times \dfrac{D}{f_e}$ Where: $m$ is the total magnification. $L$ is the tube length of the microscope. $f_o$ is the focal length of the objective lens. $D$ is the distance of distinct vision of the eye. $f_e$ is the focal length of the eyepiece.

Given: $L = 40$ cm (tube length) $f_o = 2$ cm (focal length of the objective) $D = 25$ cm (distance of distinct vision) $f_e = 4$ cm (focal length of the eyepiece) Substitute these values into the formula: $m = \dfrac{40}{2} \times \dfrac{25}{4}$ Calculate each part: $\dfrac{40}{2} = 20$ $\dfrac{25}{4} = 6.25$ Multiply these results: $m = 20 \times 6.25 = 125$ Thus, the magnification of the microscope is 125.

Q5

In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power (

p

) and magnification (

m

) for each lens will be, respectively

A

4 p

and

m^4

B

p^4

and

m^4

C

4 p

and

4 m

D

p^4

and

4 m

Correct Answer

Option A

Solution

In a camera with a series combination of four similar thin convex lenses, the effective power and total magnification of this combination can be calculated as follows: Effective Power : When lenses are in contact, their powers add up.

Therefore, the effective power $p_{\text{eff}}$ of the combination is the sum of the powers of each individual lens: $p_{\text{eff}} = p_1 + p_2 + p_3 + p_4 = 4p$ Total Magnification : The total magnification $m_{\text{eff}}$ is the product of the magnifications of each lens in the series: $m_{\text{eff}} = m_1 \times m_2 \times m_3 \times m_4 = m^4$ Thus, for a combination of four identical lenses, the effective power is four times the power of a single lens, and the effective magnification is the fourth power of the magnification of a single lens.

Q6

A light ray enters through a right angled prism at point

P

with the angle of incidence

30^{\circ}

as shown in figure. It travels through the prism parallel to its base

B C

and emerges along the face

A C

. The refractive index of the prism is:

A

\dfrac{\sqrt{5}}{4}

B

\dfrac{\sqrt{5}}{2}

C

\dfrac{\sqrt{3}}{4}

D

\dfrac{\sqrt{3}}{2}

Correct Answer

Option B

Solution

In prism,

r_1+c=A

\begin{aligned} \quad & r_1=90^{\circ}-c \quad \text{.... (1)}\\ & \sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu} \end{aligned}

$\Rightarrow$ Apply Snell's law, on incidence surface

\begin{aligned} 1 \cdot \sin 30^{\circ}=\mu \sin \left(r_1\right) \Rightarrow 1 \times \frac{1}{2}=\mu \times \sin \left(90^{\circ}-c\right) \\ \frac{1}{2}=\mu \times \frac{\sqrt{\mu^2-1}}{\mu} \end{aligned}

On squaring

\frac{1}{4}=\mu^2-1

\Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}

Q7

A small telescope has an objective of focal length

140 \mathrm{~cm}

and an eye piece of focal length

5.0 \mathrm{~cm}

. The magnifying power of telescope for viewing a distant object is:

A 34

B 28

C 17

D 32

Correct Answer

Option B

Solution

The magnifying power of a telescope when viewing distant objects can be calculated using the formula:

\text{Magnifying Power} (M) = \frac{\text{Focal Length of Objective Lens} (f_o)}{\text{Focal Length of Eyepiece Lens} (f_e)}

Given in the problem: Focal Length of Objective Lens, $f_o = 140 \mathrm{~cm}$ Focal Length of Eyepiece Lens, $f_e = 5.0 \mathrm{~cm}$ Substituting these values into the formula gives:

M = \frac{140 \mathrm{~cm}}{5.0 \mathrm{~cm}} = 28

Thus, the magnifying power of the telescope is 28 . The correct answer is Option B .

Q8

A lens is made up of 3 different transparent media as shown in figure. A point object O is placed on its axis beyond

2f

. How many real images will be obtained on the other side?

A 2

B 1

C No image will be formed

D 3

Correct Answer

Option D

Solution

Since lens is made of three materials so three $\mu$ and hence three images.

Q9

\epsilon_0

and

\mu_0

are the electric permittivity and magnetic permeability of free space respectively. If the corresponding quantities of a medium are

2 ~\epsilon_0

and

1.5 ~\mu_0

respectively, the refractive index of the medium will nearly be :

A

\sqrt{2}

B

\sqrt{3}

C 3

D 2

Correct Answer

Option B

Solution

\mu=\frac{C}{\mathrm{~V}}=\frac{\frac{1}{\sqrt{\mu_0 \in_0}}}{\frac{1}{\sqrt{1.5 \mu_0 \times 2 \epsilon_0}}}=\sqrt{3}

Q10

A horizontal ray of light is incident on the right angled prism with prism angle

6^{\circ}

. If the refractive index of the material of the prism is 1.5 , then the angle of emergence will be:

A

9^{\circ}

B

10^{\circ}

C

4^{\circ}

D

6^{\circ}

Correct Answer

Option A

Solution

A = {r_1} + {r_2} = 6^\circ

\mu = 1.5{{\sin e} \over {\sin {r_2}}} = {{\sin e} \over {\sin 6}}

\sin e = 1.5 \times 6^\circ

e \simeq 9^\circ