Ray Optics - NEET Physics Chapterwise MCQs & PYQs

NEET Ray Optics MCQs & PYQs

Practice NEET Ray Optics Questions

Question 191:

moderate

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The mass density of an optically denser medium is greater than that of an optically rarer medium.

Reason R: Optical density is found by the ratio of the speed of light in two media while mass density is mass per unit volume.

In the light of the above statements, choose the correct answer from the option given below.

Mass density and optical density are not directly related. For example, oil is optically denser than water but has a lower mass density. Thus, Assertion A is false but Reason R is true.

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Question 192:

difficult

In a compound microscope, the focal length of two lenses are \(1.25 \text{cm}\) and \(12.5 \text{cm}\). If an object is placed at \(5 \text{cm}\) from objective lens and final image is formed at \(25 \text{cm}\) from eye piece lens, the distance between the two lenses is

$$\begin{aligned} &\text{For the objective lens: } \frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \\ &\frac{1}{v_o} - \frac{1}{-5} = \frac{1}{1.25} \implies v_o = 1.67 \text{ cm} \\ &\text{For the eyepiece lens: } \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \\ &\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{12.5} \implies u_e = -8.33 \text{ cm} \\ &\text{Total distance } L = \vert{}v_o\vert{} + \vert{}u_e\vert{} = 1.67 + 8.33 = 10 \text{ cm} \end{aligned}$$
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Question 193:

easy

Given below are two statements:

Statement-I: If a parallel paraxial beam of light were incident on a concave mirror, making some angle with the principal axis, the reflected rays would converge at a point in focal plane.

Statement-II: Power(P) of concave mirror is positive even though focal length(f) is negative.

In the light of the above statements, choose the most appropriate answer from the options given below.

Statement I is a standard result of paraxial ray optics. For Statement II, power of a mirror is given by \(P = -1/f\). Since \(f\) is negative for a concave mirror, \(P\) is positive. Thus, both statements are correct.

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Question 194:

easy

A convex mirror of focal length \(10\text{ cm}\) forms an image which is \(\frac{1}{3}\) times of the height of a real object. The distance of the object from the mirror is

For a convex mirror, \(f = +10\text{ cm}\) and magnification is virtual and erect, so \(m = +\frac{1}{3}\). Using \(m = \frac{f}{f-u}\), we get \(\frac{1}{3} = \frac{10}{10-u} \implies 10-u = 30 \implies u = -20\text{ cm}\). The distance is \(20\text{ cm}\).

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Question 195:

moderate

Match the elements of List-I with List-II:

$$\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \hline \text{(A) Simple microscope} & \text{(H) Image virtual, erect and enlarged} \\ \text{(B) Compound microscope} & \text{(E) Image magnified, inverted and virtual} \\ \text{(C) Astronomical telescope} & \text{(G) Virtual, inverted and high resolution} \\ \text{(D) Terrestrial telescope} & \text{(F) Image virtual, erect and high resolution} \\ \end{array}$$

A simple microscope forms an erect, virtual, and enlarged image (A-H). A compound microscope forms a magnified, inverted, and virtual final image (B-E). An astronomical telescope has a virtual, inverted, high resolution final image (C-G). A terrestrial telescope has a virtual, erect, high resolution final image (D-F).

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Question 196:

moderate

If a container of height \(17.3\text{ cm}\) is filled with a liquid of refractive index \(\mu\). The bottom of container appear to be raised by \(3.46\text{ cm}\) when seen from above. Refractive index of liquid is

The apparent shift is \(s = d\left(1 - \frac{1}{\mu}\right)\). Substituting the values: \(3.46 = 17.3 \left(1 - \frac{1}{\mu}\right) \implies 1 - \frac{1}{\mu} = 0.2 \implies \mu = 1.25\).

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Question 197:

moderate

The refracting angle and minimum angle of deviation of a prism is same as \(60^\circ\). The refractive index of prism is

Using the prism formula: \(\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin(A/2)}\). Since \(A = D_m = 60^\circ\), \(\mu = \frac{\sin(60^\circ)}{\sin(30^\circ)} = \sqrt{3} \approx 1.732\).

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Question 198:

moderate

A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

(2010 Pre)

Focal length of a lens is independent of its aperture, so it remains $f$ . Intensity is proportional to the exposed area. Original area $A = \frac{\pi d^2}{4}$ . Covered area $a = \frac{\pi (d/2)^2}{4} = \frac{A}{4}$ . Remaining area $A' = A - \frac{A}{4} = \frac{3A}{4}$ . New intensity is proportional to remaining area, so $I' = \frac{3I}{4}$ .

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Question 199:

easy

An object is placed on the principal axis of a concave mirror at a distance of 1.5 f (f is the focal length). The image will be at,

(2020-Covid)

Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ . For a concave mirror, object distance $u = -1.5f$ and focal length is $-f$ . Substituting these values gives $\frac{1}{v} + \frac{1}{-1.5f} = \frac{1}{-f}$ , which implies $\frac{1}{v} = \frac{1}{1.5f} - \frac{1}{f} = \frac{1 - 1.5}{1.5f} = \frac{-0.5}{1.5f} = \frac{-1}{3f}$ . Thus, $v = -3f$ .

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Question 200:

moderate

An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be:

(2018)

Initially $u_1 = -40 cm$ , $f = -15 cm$ . Using mirror formula, $v_1 = \frac{u_1 f}{u_1 - f} = \frac{(-40)(-15)}{-40 - (-15)} = \frac{600}{-25} = -24 cm$ . After moving 20 cm towards the mirror, $u_2 = -20 cm$ . The new image position is $v_2 = \frac{(-20)(-15)}{-20 - (-15)} = \frac{300}{-5} = -60 cm$ . Displacement is $v_2 - v_1 = -60 - (-24) = -36 cm$ , meaning 36 cm away from the mirror.

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