Light – Reflection and Refraction — NCERT Solutions, AI Tutor & Practice

Understanding: A concave mirror reflects rays that travel parallel to its principal axis.

Definition: The principal focus of a concave mirror is the point on the principal axis at which rays of light, coming parallel to the principal axis, actually meet (converge) after reflection from the mirror.

Since the reflected rays really cross at this point, the principal focus of a concave mirror is a real point, located in front of the mirror.

Formula: For a spherical mirror, the focal length is half the radius of curvature: $f = \frac{R}{2}$

Given: $R = 24\ \text{cm}$

Substitution: $f = \frac{24}{2} = 12\ \text{cm}$

Answer: The focal length is $12\ \text{cm}$.

Analysis: A convex mirror always forms images that are:

1. Virtual, erect and diminished — the image stays upright, so the driver instantly recognises the traffic behind.
2. Reduced in size — because the image is smaller, a convex mirror covers a wider field of view than a plane mirror of the same size.

Conclusion: These two properties allow the driver to see a large area of the road behind the vehicle in an upright image, which is exactly why convex mirrors are used as rear-view mirrors.

Sign convention: Distances measured against the incident light are negative.

$u = -10\ \text{cm}$, $f = -15\ \text{cm}$

Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-10}$ $\frac{1}{v} = -\frac{1}{15} + \frac{1}{10} = \frac{-2 + 3}{30} = \frac{1}{30}$ $v = +30\ \text{cm}$

Nature: Since $v$ is positive, the image is behind the mirror, so it is virtual and erect.

Magnification: $m = -\frac{v}{u} = -\frac{30}{-10} = +3$

The image is magnified 3 times, virtual and erect — as expected when the object lies between the pole and focus of a concave mirror.

Laws of refraction:

1. The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
2. For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant (Snell's law): $\frac{\sin i}{\sin r} = \text{constant} = n_{21}$

Refractive index: The refractive index of medium 2 with respect to medium 1 is the ratio of the speed of light in medium 1 to that in medium 2: $n_{21} = \frac{v_1}{v_2}$

The absolute refractive index of a medium compares the speed of light in vacuum ($c$) with its speed in that medium: $n = \frac{c}{v}$

Formula: $n = \frac{c}{v} \quad\Rightarrow\quad v = \frac{c}{n}$

Given: $n = 1.5$, $c = 3\times10^8\ \text{m/s}$

Substitution: $v = \frac{3\times10^8}{1.5} = 2\times10^8\ \text{m/s}$

Answer: The speed of light in glass is $2\times10^8\ \text{m/s}$.

Key idea: A real image equal in size to the object is formed by a convex lens only when the object is at twice the focal length ($2f$).

Given: $f = 20\ \text{cm}$, so $2f = 40\ \text{cm}$.

Verification using the lens formula: with $u = -40\ \text{cm}$, $f = +20\ \text{cm}$: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} = \frac{1}{20} + \frac{1}{-40} = \frac{2-1}{40} = \frac{1}{40}$ $v = +40\ \text{cm}$

Magnification: $m = \dfrac{v}{u} = \dfrac{40}{-40} = -1$ (same size, inverted, real).

Answer: The object must be placed at 40 cm from the lens.

Definition: The power of a lens is the degree to which it converges or diverges light rays; it is the reciprocal of the focal length (in metres): $P = \frac{1}{f\ (\text{in m})}$ The SI unit is the dioptre (D), where $1\ \text{D} = 1\ \text{m}^{-1}$.

Given: $P = -2.5\ \text{D}$

Focal length: $f = \frac{1}{P} = \frac{1}{-2.5} = -0.4\ \text{m} = -40\ \text{cm}$

Type of lens: The negative power (and negative focal length) shows this is a concave (diverging) lens.

Given: $u = -25\ \text{cm}$, $f = +10\ \text{cm}$, object height $h = +5\ \text{cm}$.

Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} = \frac{1}{10} + \frac{1}{-25} = \frac{5 - 2}{50} = \frac{3}{50}$ $v = \frac{50}{3} \approx 16.7\ \text{cm}$

The positive value means the image is on the opposite side of the lens → real and inverted.

Magnification: $m = \frac{v}{u} = \frac{16.7}{-25} = -0.667$

Image height: $h' = m \times h = -0.667 \times 5 \approx -3.3\ \text{cm}$

Answer: The image forms about $16.7\ \text{cm}$ from the lens on the other side, is real, inverted and about $3.3\ \text{cm}$ tall (diminished).

Understanding the setup: For a concave mirror, $f$ is negative. Take an object beyond the focus, so $|u| > |f|$, with $u$ negative.

Analysis using the mirror formula: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ Both $f$ and $u$ are negative. When $|u| > |f|$, the term $\frac{1}{f}$ (a larger negative magnitude) dominates, and computing $\frac{1}{v}$ gives a negative value.

A negative $v$ means the image is formed in front of the mirror, where reflected rays actually meet.

Conclusion: Since the reflected rays genuinely converge, the image is real and inverted, not virtual. A virtual image (positive $v$) only appears when the object lies between the pole and the focus, i.e. $|u| < |f|$.

Understanding: The optical centre is the central point of a thin lens on the principal axis.

Analysis: Very close to the optical centre, the two surfaces of a thin lens are effectively parallel, like a thin flat glass slab. When light passes through a parallel-sided slab, it emerges parallel to its original direction, with only a tiny (negligible for a thin lens) sideways shift.

Conclusion: A ray of light passing through the optical centre of a thin lens goes straight through, undeviated, without any bending. This is why the ray through the optical centre is one of the standard rays used in ray diagrams.

Given: $u = -10\ \text{cm}$; a real image is inverted, so $m = -3$.

Step 1 – Find $v$ from magnification: $m = -\frac{v}{u} \Rightarrow -3 = -\frac{v}{-10}$ $-3 = -\frac{v}{-10} = \frac{v}{10}\cdot(-1)\Rightarrow v = -30\ \text{cm}$

So the image is $30\ \text{cm}$ in front of the mirror (real, inverted).

Step 2 – Find $f$ using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-30} + \frac{1}{-10}$ $\frac{1}{f} = -\frac{1}{30} - \frac{3}{30} = -\frac{4}{30} = -\frac{2}{15}$ $f = -\frac{15}{2} = -7.5\ \text{cm}$

Answer: The image is $30\ \text{cm}$ in front of the mirror and the focal length is $7.5\ \text{cm}$ (concave).