Additional ISC-Specific Numericals

PHYExam Practice (Mixed Numericals)

This is a whole-chapter numerical revision set spanning both ray and wave optics: refraction at plane surfaces (apparent depth, the critical-circle of emerging light), prisms and dispersion, lens and mirror combinations, optical instruments, interference (Young's double slit) and resolving power. The core skill examined is choosing the one correct formula for each situation and feeding it numbers with the right sign convention and consistent units.

ISC favours these because a single problem (e.g. a lens facing a mirror, or an achromatic prism combination) tests several sub-topics at once, rewarding students who can link the mirror/lens formulae, $\mu$ -based refraction relations, fringe-width and instrument magnification into one clean calculation.

Real / apparent depth (normal viewing)

\mu = \dfrac{\text{real depth}}{\text{apparent depth}} = \dfrac{h}{h'}, \qquad \text{shift} = h\left(1 - \dfrac{1}{\mu}\right)

\mu

is the refractive index of the denser medium relative to air,

h

the real depth (or slab thickness),

h'

the apparent depth seen from above. Apparent depth

h' = h/\mu

and the object appears raised by the shift

h(1-1/\mu)

. Used for the tank, coin, microscope-refocus and glass-slab problems.

Critical angle and TIR (bright circle of emerging light)

\sin C = \dfrac{1}{\mu}, \qquad r = h\tan C = \dfrac{h}{\sqrt{\mu^{2} - 1}}

C

is the critical angle for the denser-to-rarer interface,

\mu

the refractive index of the liquid (relative to air),

h

the depth of the source, and

r

the radius of the circle on the surface through which light escapes. Rays striking the surface at angles greater than

C

undergo total internal reflection. The

1/\sqrt{\mu^{2}-1}

form follows because

\tan C = 1/\sqrt{\mu^{2}-1}

Prism: minimum deviation and thin-prism deviation

\mu = \dfrac{\sin\!\left(\dfrac{A + \delta_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}, \qquad \delta \approx (\mu - 1)A \ \text{(thin prism)}

A

is the refracting angle,

\delta_m

the angle of minimum deviation,

\mu

the refractive index of the prism material relative to its surroundings. For a prism immersed in a medium use the relative index

{}_{m}\mu_g = \mu_g/\mu_m

, which is closer to

1

and so reduces

\delta_m

. The small-angle form

\delta \approx (\mu-1)A

holds only for thin prisms, not for a

60^{\circ}

prism.

Lens maker's formula and lens / mirror formulae

\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right), \quad \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\ \text{(lens)}, \quad \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\ \text{(mirror)}

f

is focal length,

\mu

the lens material's refractive index relative to its surroundings,

R_1, R_2

the radii of curvature (Cartesian sign convention; for an equiconvex lens

R_1 = +R

R_2 = -R

u, v

are object/image distances;

R = 2f

for a spherical mirror. The lens and mirror formulae chain together when the image from one acts as the object for the next.

Magnifying power of microscope and telescope

M_{\text{simple}} = 1 + \dfrac{D}{f}\ (\text{near pt}),\ \dfrac{D}{f}\ (\infty); \qquad M_{\text{tele}} = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right),\ L = f_o + f_e

D = 25\,\text{cm}

is the least distance of distinct vision,

f

the simple-lens focal length,

f_o, f_e

the objective and eyepiece focal lengths. The bracket

(1 + f_e/D)

applies for the final image at the near point; for the image at infinity (normal adjustment)

M = f_o/f_e

and the tube length is

L = f_o + f_e

Young's double slit: fringe width and maxima

\beta = \dfrac{\lambda D}{d}, \qquad y_n^{\text{bright}} = \dfrac{n\lambda D}{d}, \qquad \Delta x = d\sin\theta = n\lambda

\beta

is the fringe width (spacing between adjacent bright or dark fringes),

\lambda

the wavelength,

d

the slit separation,

D

the slit-to-screen distance, and

y_n

the position of the

n

th bright fringe from the centre. The path difference for a bright fringe is

\Delta x = n\lambda

(

n = 0, 1, 2,\dots

); for a dark fringe it is

(n - \tfrac{1}{2})\lambda

. Keep

\lambda, d, D

in consistent units.

Limit of resolution (Rayleigh, circular aperture)

\Delta\theta = \dfrac{1.22\,\lambda}{D}

\Delta\theta

is the smallest resolvable angular separation (radians),

\lambda

the wavelength of light,

D

the aperture (objective) diameter. The resolving power is

1/\Delta\theta

; a larger aperture or shorter wavelength resolves finer detail.

Sign convention is non-negotiable: measure all distances from the pole/optic centre, take the incident-light direction as positive. With this, a real object gives $u$ negative and a convex mirror has $f$ positive (concave mirror $f$ negative); plug signed values in and let the formula return the sign of $v$ .
Apparent-depth formula $\mu = h/h'$ is for near-normal (paraxial) viewing only. The object is raised by $h(1 - 1/\mu)$ , which depends only on thickness and $\mu$ , never on how far above the surface your eye sits.
For the 'bright circle' problem, light escapes only inside the critical cone: any ray hitting the surface at more than the critical angle $C$ is totally internally reflected, so the lit patch has radius $r = h\tan C$ with $\sin C = 1/\mu$ .
A prism's deviation depends on its index relative to the surroundings: moving glass ( $\mu_g$ ) from air into water replaces $\mu_g$ by ${}_{m}\mu_g = \mu_g/\mu_m$ (so $\delta_m$ drops sharply), and use $\delta \approx (\mu - 1)A$ only for thin prisms, never for a $60^{\circ}$ prism.
Dispersion uses deviation $\delta = (\mu - 1)A$ per prism and angular dispersion $= \omega\delta$ . For deviation without dispersion (achromatic prism) the two prisms are oppositely oriented and you set $\omega_1\delta_1 + \omega_2\delta_2 = 0$ , leaving a net mean deviation $\delta_1 - \delta_2$ ; for dispersion without deviation (direct-vision) you instead set $\delta_1 - \delta_2 = 0$ .
In a lens–mirror combination where the final image coincides with the object, rays must retrace their path and so strike a concave (spherical) mirror normally; this means the lens image lands at the mirror's centre of curvature, so trace the lens image first, then use $R = 2f$ for the mirror.
For a glass-in-air equiconvex lens, $R_1 = +R$ and $R_2 = -R$ , so $1/f = (\mu - 1)(2/R)$ , giving $f = R/[2(\mu - 1)]$ ; with $\mu = 1.5$ this conveniently makes $f = R$ .
Match the magnification formula to the eye's focus: $M = D/f$ (or $f_o/f_e$ ) when the final image is at infinity (relaxed eye), and the larger near-point form when the image is at $D = 25\,\text{cm}$ . Fringe width $\beta = \lambda D/d$ scales directly with $\lambda$ and $D$ and inversely with $d$ .

Working a problem that spans two optical elements

1Treat the elements one at a time, in the order the light meets them.
2The image formed by the first element is the object for the second — carry its position across, and keep the sign convention anchored at each element's own pole or optic centre.
3A virtual image from the first element becomes a virtual object for the second, so its distance changes sign; this is where most marks are lost.
4When the final image coincides with the object, the light must be retracing its path, which means it struck a mirror normally — so the rays hit the mirror's centre of curvature.
5Only convert to a single equivalent element (for example a silvered lens as an equivalent mirror) once you can state the combined focal length; otherwise work element by element.

Where the marks go

Confusing real and apparent depth: students divide by $\mu$ to get the raise instead of the apparent depth. Apparent depth is $h/\mu$ ; the amount raised is $h - h/\mu = h(1 - 1/\mu)$ — two different numbers.
Using $\sin C$ where $\tan C$ is needed: the critical angle comes from $\sin C = 1/\mu$ , but the circle's radius needs $r = h\tan C$ . Writing $r = h\sin C$ is a classic error.
Mishandling radius signs in the lens maker's formula (using $1/R_1 + 1/R_2$ , or treating both radii as positive), which flips a converging lens to diverging or doubles the focal length; keep $R_1 = +R$ , $R_2 = -R$ for an equiconvex lens.
In interference, mixing units of $\lambda, d, D$ (e.g. $\lambda$ in nm but $d$ in mm) or using $\theta$ instead of $\sin\theta$ in the path difference; for small angles $\beta = \lambda D/d$ only works when all lengths share one unit and the answer for $\beta$ comes out as a length.

How the board asks it

Numericallens and mirror formulae with sign convention
A convex lens of focal length $20\,\text{cm}$ is placed coaxially $5\,\text{cm}$ in front of a concave mirror of radius of curvature $30\,\text{cm}$ . An object is kept $30\,\text{cm}$ in front of the lens. Calculate the position and nature of the final image formed by the lens-mirror combination.
Numericalyoung's double slit fringe width
In a Young's double-slit experiment, two slits $0.2\,\text{mm}$ apart are illuminated by light of wavelength $600\,\text{nm}$ , with the screen $1.5\,\text{m}$ away. Calculate the fringe width and the distance of the $4^{\text{th}}$ bright fringe from the central maximum.
Numericaltotal internal reflection and the critical-circle of emerging light
A point source of light lies at the bottom of a tank of water ( $\mu = 4/3$ ) of depth $1\,\text{m}$ . Calculate the critical angle and hence the radius of the bright circle of light through which light emerges at the water surface.
Numericalthin-prism deviation in a surrounding medium
A thin prism of refracting angle $A = 6^{\circ}$ made of glass ( $\mu_g = 1.5$ ) is immersed in water ( $\mu_w = 4/3$ ). Calculate its angle of deviation in water and compare it with the deviation it produces in air.
Give reasonsrays retracing their path onto a mirror's centre of curvature
An object is placed in front of a convex lens with a concave mirror behind it, and the final image is found to coincide with the object itself. State, with reasoning, where the image formed by the lens alone must lie, and outline the steps to find the focal length of the mirror.
Derive / provediffraction limit and resolving power of a telescope
Obtain an expression for the limit of resolution of a telescope objective of aperture $D$ for light of wavelength $\lambda$ , and calculate it for $D = 10\,\text{cm}$ and $\lambda = 500\,\text{nm}$ .

PreviousOptical Instruments (Microscope & Telescope)

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.