Ray Optics and Optical Instruments: NEET previous year questions

When the ray inside an equilateral prism ($A = 60^{\circ}$) travels parallel to the base, the prism is in the minimum-deviation condition where the angle of incidence equals the angle of emergence ($i = e$). Using $i + e = A + \delta$ gives $\delta = 2i - A = 2(50^{\circ}) - 60^{\circ} = 40^{\circ}$, option (3).

A concave lens is diverging, so a ray incident parallel to the principal axis bends away from the axis and travels as though it originated from the focal point on the same (incident) side of the lens. Traced backwards, the emergent ray appears to diverge from that near-side focus. Strictly, this near-side point is the second principal focus ($F_2$ in the figure), but option (2) is still the best answer, because it is the only option where the ray appears to diverge from a focus.

Why the other options are wrong:

- (1) A parallel ray relates to the focus, not to $2F$; this describes a different ray.
- (3) It does not stay parallel; a diverging lens spreads a parallel beam.
- (4) The emergent ray only appears to come from the second principal focus $F_2$ on the incident side; it moves away from that point and never passes through it.

On an optical bench the positions read on the scale are those of the UPRIGHTS, not of the optical centres of the lens, mirror or object pin. The fixed offset between an upright's index mark and the actual optical element is removed by the index (or bench) correction, applied to every measured object and image distance before computing the focal length. NCERT Class 12 practical work, Ray Optics and Optical Instruments.

For near-normal viewing across a plane boundary, apparent depth $=$ real depth $\times \frac{n_{\text{observer}}}{n_{\text{object}}}$, and the object sits 2.5 cm from each face. Viewed from $P$: $h_1 = 2.5 \times \frac{1}{1.25} = 2\ \text{cm}$. Viewed from $R$: $h_2 = 2.5 \times \frac{1.5}{1.25} = 3\ \text{cm}$. Hence $|h_1 - h_2| = 1\ \text{cm}$. Looking in from the DENSER side makes the object appear farther away, which is the sense check here. NCERT Class 12, Ray Optics and Optical Instruments (refraction at plane surfaces).

Why the other options are wrong:

- (1) assumes the symmetric placement of the object makes the two apparent depths equal, ignoring the different observer media
- (3) uses the full 5 cm thickness instead of the 2.5 cm from the object to each face
- (4) computes $h_2$ alone and reports it as the difference

For the convex lens, $u = -30\ \text{cm}$ and $f = +10\ \text{cm}$ give $\frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{1}{15}$, so its image lies 15 cm to its right. The lenses are 3 cm apart, so that image sits 12 cm BEYOND the concave lens and acts as a VIRTUAL OBJECT with $u = +12\ \text{cm}$. Then $\frac{1}{v} = -\frac{1}{10} + \frac{1}{12} = -\frac{1}{60}$, giving $v = -60\ \text{cm}$: 60 cm to the left of the concave lens. NCERT Class 12, Ray Optics and Optical Instruments (combination of thin lenses).

Why the other options are wrong:

- (1) uses the thin-lens-in-contact formula $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$, which fails for a 3 cm separation
- (3) keeps the sign of $v$ positive, missing that a negative $v$ places a virtual image on the incoming side
- (4) gets the magnitude right but assigns the image to the wrong side of the concave lens

Because the ray emerges normally, the second internal angle is zero and the first is approximately the prism angle A. For a small angle, Snell's law gives i $\approx$ $\mu$A. Chapter: Ray Optics and Optical Instruments.

The limit of resolution $\theta$ for a telescope is given by $\theta = 1.22 \frac{\lambda}{D}$. Substituting the values, $\theta = 1.22 \frac{600 \times 10^{-9}}{2} = 3.66 \times 10^{-7}\ rad$. Chapter: Ray Optics and Optical Instruments.

Brewster's angle $i_b$ is given by $\tan i_b = \mu$, where $\mu$ is the refractive index of the second medium with respect to the first. Since $\mu > 1$, $i_b$ must be greater than 45° but less than 90°. Chapter: Ray Optics and Optical Instruments.