IMO Practice Test: Optics | Class 12 Physics

Question 1 of 14

A concave mirror of focal length $12\,\text{cm}$ forms a real image at $24\,\text{cm}$. The object distance is:

$8\,\text{cm}$

$24\,\text{cm}$

$12\,\text{cm}$

$36\,\text{cm}$

Explanation: $1/u = 1/f - 1/v = -1/12 + 1/(-24)$... using $1/v+1/u=1/f$ with $v=-24,f=-12$: $1/u=-1/12+1/24=-1/24 \Rightarrow u=-24\,\text{cm}$.

Question 2 of 14

If the object distance equals the focal length of a convex lens, the image forms at:

the focus

the optical centre

$2f$

infinity

Explanation: For $u = f$, rays emerge parallel and the image is at infinity.

Question 3 of 14

A ray hits a glass slab ($n=1.5$) and emerges from the parallel face. The emergent ray is:

bent towards normal

parallel to incident ray but laterally shifted

reflected back

totally internally reflected

Explanation: A parallel-sided slab produces only a lateral shift; the emergent ray is parallel to the incident ray.

Question 4 of 14

Two thin lenses of focal lengths $+20\,\text{cm}$ and $-30\,\text{cm}$ in contact form a combination of focal length:

$+12\,\text{cm}$

$-60\,\text{cm}$

$+50\,\text{cm}$

$+60\,\text{cm}$

Explanation: $1/f = 1/20 - 1/30 = (3-2)/60 = 1/60 \Rightarrow f = +60\,\text{cm}$.

Question 5 of 14

The refractive index of a prism is $\sqrt{2}$ and refracting angle $60^\circ$. The angle of minimum deviation is:

$30^\circ$

$45^\circ$

$60^\circ$

$90^\circ$

Explanation: $\sin\frac{A+\delta_m}{2} = \sqrt{2}\sin 30^\circ = \sqrt{2}/2 = \sin 45^\circ \Rightarrow (A+\delta_m)/2 = 45 \Rightarrow \delta_m = 30^\circ$.

Question 6 of 14

In Young\u2019s experiment, if the whole apparatus is immersed in water ($n=4/3$), the fringe width:

increases

is unchanged

becomes zero

decreases

Explanation: Wavelength reduces to $\lambda/n$, so $\beta = \lambda D/(nd)$ decreases.

Question 7 of 14

The path difference for the $3^{rd}$ dark fringe from the centre is:

$3\lambda$

$2.5\lambda$

$3.5\lambda$

$2\lambda$

Explanation: Dark fringes: $\Delta = (n+\tfrac{1}{2})\lambda$; the 3rd dark fringe has $n=2$, giving $2.5\lambda$.

Question 8 of 14

Polarised light of intensity $I_0$ becomes $I_0/2$ after an analyser when $\theta$ equals:

$0^\circ$

$30^\circ$

$60^\circ$

$45^\circ$

Explanation: $\cos^2\theta = 1/2 \Rightarrow \theta = 45^\circ$.

Question 9 of 14

A point object is placed at the centre of curvature of a concave mirror. The image coincides with the object because:

rays retrace their path

the focus lies at C

the mirror is plane

no reflection occurs

Explanation: Rays through $C$ strike the mirror normally and retrace, forming the image at $C$ itself.

Question 10 of 14

Light of wavelength $500\,\text{nm}$ gives a first diffraction minimum at $30^\circ$ for a slit of width:

$0.5\,\mu\text{m}$

$1\,\mu\text{m}$

$2\,\mu\text{m}$

$5\,\mu\text{m}$

Explanation: $a = \lambda/\sin\theta = 500\times10^{-9}/0.5 = 1\times10^{-6}\,\text{m} = 1\,\mu\text{m}$.

Question 11 of 14

For a glass-air interface, Brewster\u2019s angle satisfies:

$\sin i_B = n$

$\cos i_B = n$

$\tan i_B = n$

$\tan i_B = 1/n$

Explanation: At Brewster\u2019s angle $\tan i_B = n$; the reflected ray is fully polarised.

Question 12 of 14

A convex lens of focal length $f$ produces a same-size real image when the object is at:

$f$

$2f$

infinity

$f/2$

Explanation: At $u = 2f$, $v = 2f$ and $|m| = 1$ (real, inverted, same size).

Question 13 of 14

When a concave mirror is half-covered with an opaque card, the image:

becomes half

disappears

is complete but dimmer

is inverted twice

Explanation: Each part of the mirror forms a full image; covering half only reduces brightness.

Question 14 of 14

Two coherent sources have intensities in ratio $1:4$. The ratio of maximum to minimum intensity in interference is:

$9:1$

$5:3$

$3:1$

$25:9$

Explanation: Amplitudes ratio $1:2$; $I_{max}/I_{min} = (2+1)^2/(2-1)^2 = 9/1$.

IMO Practice Test — Optics