NEET Ray Optics MCQs & PYQs

Q: A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is: (2012 Mains)

The rod's closer end is at $u_1 = -20 cm$ (which is $2f$ since $f = -10 cm$ ). The image of this end forms at the center of curvature, $v_1 = -20 cm$ . The farther end is at $u_2 = -30 cm$ . Its image is at $v_2 = \frac{u_2 f}{u_2 - f} = \frac{(-30)(-10)}{-30 - (-10)} = \frac{300}{-20} = -15 cm$ . The length of the image is $|v_1 - v_2| = |-20 - (-15)| = 5 cm$ .

Q: A concave mirror of focal length $f_1$ is placed at a distance of d from a convex lens of focal length $f_2$ . A beam of light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance 'd' must equal: (2012 Pre)

Light from infinity passes through the convex lens and converges at its principal focus, a distance $f_2$ from the lens. For the light to return to infinity, it must retrace its path after reflection from the concave mirror. This happens if the rays strike the mirror normally, meaning they appear to come from the mirror's center of curvature (at distance $2f_1$ ). So, $d = 2f_1 + f_2$ .

Q: A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirror will be: (2000)

For a person to see their full, complete image in a vertical plane mirror, the minimum length of the mirror required is exactly half of the person's height, regardless of their distance from the mirror. Therefore, minimum length = $6 / 2 = 3 feet$ .

Q: The frequency of a light wave in a material is $2 \times 10^{14} Hz$ and wavelength is $5000\mathring{A}$ . The refractive index of material will be: (2007)

Velocity of light in the material is given by $v = f \lambda = (2 \times 10^{14} Hz) \times (5000 \times 10^{-10} m) = 10^8 m/s$ . The refractive index of the material is calculated as $n = \frac{c}{v} = \frac{3 \times 10^8}{10^8} = 3.00$ .

Q: A ray of light travelling in air haves wavelength $\lambda$ , frequency n, velocity v and intensity, I. If this ray enters into water than these parameter are $\lambda'$ , n', v' and I' respectively. Which relation is correct: (2001)

When a light ray travels from one medium to another, its frequency remains absolutely unchanged because frequency is a fundamental characteristic of the source. Velocity and wavelength change based on the refractive index. Thus, $n = n'$ .

Q: The refractive index of water is 1.33. What will be the speed of light in water? (1996)

The speed of light in a given medium is related to the refractive index by $v = \frac{c}{n}$ . Given the speed of light in vacuum $c = 3 \times 10^8 m/s$ and the refractive index of water $n = 1.33 = \frac{4}{3}$ . We find $v = \frac{3 \times 10^8}{4/3} = \frac{9 \times 10^8}{4} = 2.25 \times 10^8 m/s$ .

Q: Time taken by sunlight to pass through a window of thickness $4 mm$ whose refractive index is $\frac{3}{2}$ is (1993)

Velocity of light in the window glass is $v = \frac{c}{\mu} = \frac{3 \times 10^8}{3/2} = 2 \times 10^8 m/s$. Thickness is $t = 4 mm = 4 \times 10^{-3} m$. Time taken is $T = \frac{t}{v} = \frac{4 \times 10^{-3}}{2 \times 10^8} = 2 \times 10^{-11} s$.

Q: An air bubble in a glass slab with refractive index $1.5$ (near normal incidence) is $5 cm$ deep when viewed from one surface and $3 cm$ deep when viewed from the opposite surface. The thickness (in cm) of the slab is: (2016 - II)

Total apparent depth of the air bubble is the sum of apparent depths from both sides: $d_{app} = 5 + 3 = 8 cm$. Apparent depth is related to real thickness T by $d_{app} = \frac{T}{\mu}$. Therefore, real thickness $T = d_{app} \times \mu = 8 \times 1.5 = 12 cm$.

Q: A bubble in glass slab ($\mu = 1.5$) when viewed from one side appears at $5 cm$ and $2 cm$ from other side, then thickness of slab is: (2000)

Total apparent thickness of the glass slab is $d_{app} = 5 + 2 = 7 cm$. Using the relation $d_{app} = \frac{t}{\mu}$, the real thickness is $t = d_{app} \times \mu$. Substituting the given values, $t = 7 \times 1.5 = 10.5 cm$.

Question 1

Match the corresponding entries of column-1 with column-2. [where m is the magnification produced by the mirror]

\begin{array}{ll}
\textbf{Column-I} & \textbf{Column-II} \\[1ex]
\text{(A) } m = -2 & \text{(1) Convex mirror} \\
\text{(B) } m = -\frac{1}{2} & \text{(2) Concave mirror} \\
\text{(C) } m = +2 & \text{(3) Real image} \\
\text{(D) } m = +\sqrt{\frac{2}{3}} & \text{(4) Virtual image}
\end{array}

(2016 – I)

Answer

Negative magnification implies a real inverted image, which is only formed by a concave mirror. So A ( $m = -2$ ) and B ( $m = -1/2$ ) match with (2) Concave and (3) Real. Positive magnification > 1 ( $m = +2$ ) implies a virtual magnified image, formed only by a concave mirror (2, 4). Positive magnification < 1 implies a virtual diminished image, formed by a convex mirror (1, 4).

💬 Discuss this Question

Question 2

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is:

(2012 Mains)

Answer

The rod's closer end is at $u_1 = -20 cm$ (which is $2f$ since $f = -10 cm$ ). The image of this end forms at the center of curvature, $v_1 = -20 cm$ . The farther end is at $u_2 = -30 cm$ . Its image is at $v_2 = \frac{u_2 f}{u_2 - f} = \frac{(-30)(-10)}{-30 - (-10)} = \frac{300}{-20} = -15 cm$ . The length of the image is $|v_1 - v_2| = |-20 - (-15)| = 5 cm$ .

💬 Discuss this Question

Question 3

A concave mirror of focal length $f_1$ is placed at a distance of d from a convex lens of focal length $f_2$ . A beam of light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance ‘d’ must equal:

(2012 Pre)

Answer

Light from infinity passes through the convex lens and converges at its principal focus, a distance $f_2$ from the lens. For the light to return to infinity, it must retrace its path after reflection from the concave mirror. This happens if the rays strike the mirror normally, meaning they appear to come from the mirror's center of curvature (at distance $2f_1$ ). So, $d = 2f_1 + f_2$ .

💬 Discuss this Question

Question 4

A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirror will be:

(2000)

Answer

For a person to see their full, complete image in a vertical plane mirror, the minimum length of the mirror required is exactly half of the person's height, regardless of their distance from the mirror. Therefore, minimum length = $6 / 2 = 3 feet$ .

💬 Discuss this Question

Question 5

The frequency of a light wave in a material is $2 \times 10^{14} Hz$ and wavelength is $5000\mathring{A}$ . The refractive index of material will be:

(2007)

Answer

Velocity of light in the material is given by $v = f \lambda = (2 \times 10^{14} Hz) \times (5000 \times 10^{-10} m) = 10^8 m/s$ . The refractive index of the material is calculated as $n = \frac{c}{v} = \frac{3 \times 10^8}{10^8} = 3.00$ .

💬 Discuss this Question

Question 6

A ray of light travelling in air haves wavelength $\lambda$ , frequency n, velocity v and intensity, I. If this ray enters into water than these parameter are $\lambda’$ , n’, v’ and I’ respectively. Which relation is correct:

(2001)

Answer

When a light ray travels from one medium to another, its frequency remains absolutely unchanged because frequency is a fundamental characteristic of the source. Velocity and wavelength change based on the refractive index. Thus, $n = n'$ .

💬 Discuss this Question

Question 7

The refractive index of water is 1.33. What will be the speed of light in water?

(1996)

Answer

The speed of light in a given medium is related to the refractive index by $v = \frac{c}{n}$ . Given the speed of light in vacuum $c = 3 \times 10^8 m/s$ and the refractive index of water $n = 1.33 = \frac{4}{3}$ . We find $v = \frac{3 \times 10^8}{4/3} = \frac{9 \times 10^8}{4} = 2.25 \times 10^8 m/s$ .

💬 Discuss this Question

Question 8

Time taken by sunlight to pass through a window of thickness $4 mm$ whose refractive index is $\frac{3}{2}$ is

(1993)

Answer

Velocity of light in the window glass is $v = \frac{c}{\mu} = \frac{3 \times 10^8}{3/2} = 2 \times 10^8 m/s$.
Thickness is $t = 4 mm = 4 \times 10^{-3} m$.
Time taken is $T = \frac{t}{v} = \frac{4 \times 10^{-3}}{2 \times 10^8} = 2 \times 10^{-11} s$.

💬 Discuss this Question

Question 9

An air bubble in a glass slab with refractive index $1.5$ (near normal incidence) is $5 cm$ deep when viewed from one surface and $3 cm$ deep when viewed from the opposite surface. The thickness (in cm) of the slab is:

(2016 – II)

Answer

Total apparent depth of the air bubble is the sum of apparent depths from both sides: $d_{app} = 5 + 3 = 8 cm$.
Apparent depth is related to real thickness T by $d_{app} = \frac{T}{\mu}$.
Therefore, real thickness $T = d_{app} \times \mu = 8 \times 1.5 = 12 cm$.

💬 Discuss this Question

Question 10

A bubble in glass slab ($\mu = 1.5$) when viewed from one side appears at $5 cm$ and $2 cm$ from other side, then thickness of slab is: (2000)

Answer

Total apparent thickness of the glass slab is $d_{app} = 5 + 2 = 7 cm$.
Using the relation $d_{app} = \frac{t}{\mu}$, the real thickness is $t = d_{app} \times \mu$.
Substituting the given values, $t = 7 \times 1.5 = 10.5 cm$.

💬 Discuss this Question

NEET Ray Optics MCQs & PYQs

Practice NEET Ray Optics Questions