NEET Practice Questions

Q231. For a prism, when the light undergoes minimum deviation, the relationship between the angle of incidence $(i)$ and the angle of emergence $(i’)$ is:

$i=i’$
$i>i’$
$i<i’$
$i=0$

Explanation:

Hint:

i = e

$i =e$ which implies

r_{1} = r_{2}

$r_1 = r_2$ .

Step: Find the relation between the angles $i$ and $i'$ .
The angle of incidence $i$ and angle of emergence $i'$ on the prism is shown in the figure below;

For the minimum angle of deviation;
$i = i'$
$\Rightarrow r_1 = r_2$
Therefore, the correct relation between the angles is $i = i'$ .
Hence, option (1) is the correct answer.

#456714

Exam: NEET – 2024

Q232. A small telescope has an objective of focal length $140~\text{cm}$ and an eyepiece of focal length $5.0~\text{cm}.$ The magnifying power of the telescope for viewing a distant object is:

$28$
$17$
$32$
$34$

Explanation:

Hint:

m = \frac{f_{0}}{f_{e}}

$m = \frac{f_0}{f_e}$

Step: Find the magnifying power of the telescope for viewing a distant object.
Given: $f_0 = 140~\text{cm}, f_e = 5~\text{cm}$
The magnifying power of a telescope for viewing a distant object is given by;
$m = \frac{f_0}{f_e}$
$\Rightarrow m = \frac{140}{5}= 28$
Therefore, the magnifying power of the telescope is $28$ .
Hence, option (1) is the correct answer.

#454989

Exam: NEET – 2024

Q233. A light ray enters through a right-angled prism at point $P$ with the angle of incidence $30^\circ$ as shown in the figure. It travels through the prism parallel to its base $BC$ and emerges along the face $AC.$ The refractive index of the prism is:

${\dfrac{\sqrt5}{2}}$
${\dfrac{\sqrt3}{4}}$
${\dfrac{\sqrt3}{2}}$
${\dfrac{\sqrt5}{4}}$

Explanation:

Hint: Apply Snell’s law.

Step: Find the refractive index of the prism.
Given: $A= 90^{\circ}$
Let the refractive index of the prism is $\mu$ .
$A= r_1+ r_2$
$\Rightarrow r_2 = 90^{\circ} - r_1$
Apply Snell’s law at point $P$
$1\times \sin30^{\circ}= \mu \sin r_1\Rightarrow \mu \sin r_1 = \frac{1}{2}~~~...(1)$
Apply Snell’s law at point $Q$
$\mu\times \sin r_2= 1 \sin90^{\circ}\Rightarrow \mu \sin r_2= 1\Rightarrow \mu \sin{(90-r_1)}=1$
$\Rightarrow \mu \cos r_1= 1~~~...(2)$
Squaring and adding equations (1) and (2) we get;
$\mu^2 = 1+\frac{1}{4}= \frac{5}{4}$
$\Rightarrow \mu = \frac{\sqrt{5}}{2}$
Therefore, the refractive index of the prism is $\frac{\sqrt{5}}{2}$ .
Hence, option (1) is the correct answer.

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Exam: NEET – 2024

Q234. 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 –

$4 p$ and $m^4$
$p^4$ and $m^4$
$4 p$ and $4 m$
$p^4$ and $4 m$

Exam: NEET – 2025

Q235. A microscope has an objective of focal length $2~\text{cm},$ eyepiece of focal length $4~\text{cm}$ and the tube length of $40~\text{cm}.$ If the distance of distinct vision of eye is $25~\text{cm},$ the magnification in the microscope is:

$150$
$250$
$100$
$125$

Exam: NEET – 2025

Q236. A ray of monochromatic light is passing through an equilateral prism $(ABC)$ as shown in the figure. The refracted ray $(QR)$ is parallel to its base $(BC)$ and the angle of incidence $(i)$ is $50^\circ$ . Then the angle of deviation $(\delta)$ is:

$40^\circ$
$45^\circ$
$55^\circ$
$35^\circ$

Explanation:

Hint:

δ_{m i n} = i + e - A

$\delta _{\mathrm{min}}=i+e-A$

Step: Find the angle of deviation.
The deviation in the prism is given by;
$\delta = i +e-A$
Where, $A = r_1 +r_2$ , since the path is symmetric ( $(r_1 = r_2)$ ) so, the angle of incidence is equal to angle of emergence i.e., $i = e = 50^{\circ}$
Therefore, the angle of deviation is given by;
$\Rightarrow \delta = 50^{\circ} +50^{\circ}-60^{\circ} = 40^{\circ}$
Hence, option (1) is the correct answer.

#592783

Exam: NEET – 2026

Q237. In a concave lens, a ray of light emanating from the object parallel to the principal axis of the lens, after refraction:

emerges parallel to the principal axis
appears to diverge from the first principal focus
passes through $2F,$ which is the radius of curvature of the lens
passes through the second principal focus

Exam: NEET – 2026

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