Wave Optics Formula Sheet: All Key Formulas for NEET

Physics · Wave Optics · NEET

Wave Optics gives NEET 1-2 questions almost every year, and most of them just need one plugged-in formula. The must-know ones are fringe width beta = lambda*D/d, intensity I = I_max*cos^2(phi/2), single-slit minima a*sin(theta) = n*lambda, resolving power (1.22*lambda/D), Malus law I = I_0*cos^2(theta), and Brewster tan(i_B) = mu. Memory hook: "Big Fringe = Distance over slit gap" -> beta grows when D grows or d shrinks.
Wave Optics: Core NEET FormulasYDSEbeta = lambda*D/dI = Imax cos^2(phi/2)Diffractiona sin(theta) = n*lambda (dark)Z_F = a^2/lambdaResolutiond_theta = 1.22*lambda/DRP = 1/d_thetaPolarisationMalus: I = I0 cos^2(theta)Brewster: tan(i_B) = muPhase / Pathphi = (2pi/lambda) x path diffBright: n*lambda Dark: (2n-1)lambda/2
One-glance map of the Wave Optics formulas NEET repeats: YDSE fringe width and intensity, single-slit diffraction and Fresnel distance, resolving power, polarisation (Malus and Brewster), and the phase-path relation.

Your doubts, answered

What is the fringe width formula and what does each letter mean?

Fringe width beta = lambda*D/d. Here lambda is the wavelength of light, D is the distance from the slits to the screen, and d is the gap between the two slits. beta is the distance between two neighbouring bright (or dark) fringes. Bright and dark fringes have the same width. The angular fringe width is theta = beta/D = lambda/d, which does not depend on D.

When do I use a*sin(theta) = n*lambda, minima or maxima?

In single-slit diffraction, a*sin(theta) = n*lambda (with n = 1, 2, 3...) gives the DARK bands (minima). This is the opposite of YDSE, where n*lambda gives BRIGHT fringes. The secondary maxima of a single slit are roughly at a*sin(theta) = (2n+1)*lambda/2. Mixing up the single-slit and double-slit conditions is the most common NEET slip.

How are path difference and phase difference related?

Phase difference phi = (2*pi/lambda) * (path difference). So a path difference of one full lambda equals a phase difference of 2*pi (360 degrees), and lambda/2 equals pi (180 degrees). Bright fringe: path difference = n*lambda. Dark fringe: path difference = (2n-1)*lambda/2.

What are the intensity formulas for two interfering waves?

For two coherent waves, resultant intensity I = I_1 + I_2 + 2*sqrt(I_1*I_2)*cos(phi). If both waves have equal intensity I_0, this becomes I = 4*I_0*cos^2(phi/2), and I_max = 4*I_0, I_min = 0. The ratio I_max/I_min = ((sqrt(I_1)+sqrt(I_2))/(sqrt(I_1)-sqrt(I_2)))^2, or in amplitude form ((a_1+a_2)/(a_1-a_2))^2.

What is Malus's law and Brewster's law?

Malus's law: when polarised light of intensity I_0 passes through a polaroid whose axis is at angle theta, transmitted intensity I = I_0*cos^2(theta). For unpolarised light hitting the first polaroid, output is I_0/2 (average of cos^2 is 1/2). Brewster's law: at the polarising angle, tan(i_B) = mu, and the reflected and refracted rays are perpendicular (i_B + r = 90 degrees). The reflected light is fully polarised.

What is the resolving power / limit of resolution formula?

Limit of resolution (smallest angle you can just separate): d_theta = 1.22*lambda/D for a telescope, where D is the objective diameter. Resolving power = 1/d_theta = D/(1.22*lambda). For a microscope, resolving power = 2*mu*sin(beta)/(1.22*lambda) (the numerator 2*mu*sin(beta) is the numerical aperture term). Both improve when wavelength lambda is smaller.

What is the Fresnel distance formula?

Fresnel distance Z_F = a^2/lambda, where a is the aperture (slit) size. For distances less than Z_F the beam is roughly a straight ray (ray optics works); beyond Z_F diffraction spreading becomes important. This marks where ray optics breaks down.

⚠️ The NEET trap
Using a*sin(theta) = n*lambda to find BRIGHT fringes.
In single-slit diffraction a*sin(theta) = n*lambda gives DARK minima; in YDSE d*sin(theta) = n*lambda gives BRIGHT maxima. Read whether the slit is single or double before you decide bright or dark.
🧠 Single-slit vs double-slit: same lambda, opposite meaning.

Real NEET questions

2016

A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 x 10^-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

A · 0.10 cm
B · 0.25 cm
C · 0.20 cm
D · 0.15 cm
Solution: Single-slit first minimum: a*sin(theta) = lambda. On the screen y = f*tan(theta) is approximately f*lambda/a. So y = 60 * (5 x 10^-5) / 0.02 = (3 x 10^-3)/0.02 = 0.15 cm. Answer 0.15 cm.
2020

Light of wavelength 600 nm is coming from a star. The limit of resolution of a telescope whose objective has a diameter of 2 m is

A · 7.32 x 10^-7 rad
B · 6.00 x 10^-7 rad
C · 3.66 x 10^-7 rad
D · 1.83 x 10^-7 rad
Solution: Limit of resolution d_theta = 1.22*lambda/D. Here lambda = 600 x 10^-9 m and D = 2 m. d_theta = 1.22 * (600 x 10^-9) / 2 = 1.22 * 3 x 10^-7 = 3.66 x 10^-7 rad. Answer 3.66 x 10^-7 rad.
2026

In Young's double slit experiment, using monochromatic light of wavelength lambda, the intensity of light at a point where the path difference is lambda is K units. The intensity at a point where the path difference is lambda/3 will be

A · K/4
B · K
C · 2K
D · K/2
Solution: I = I_max*cos^2(phi/2) with phi = (2*pi/lambda)*(path difference). At path difference lambda, phi = 2*pi, cos^2(pi) = 1, so I = I_max = K, meaning I_max = K. At path difference lambda/3, phi = 2*pi/3, cos^2(pi/3) = (1/2)^2 = 1/4, so I = K/4. Answer K/4.

Solved Wave Optics NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

See all 22 Wave Optics NEET PYQs ›
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Frequently asked

Which wave optics formula is asked most in NEET?

YDSE formulas dominate: fringe width beta = lambda*D/d and intensity I = I_max*cos^2(phi/2). Single-slit diffraction a*sin(theta) = n*lambda, resolving power 1.22*lambda/D, Malus law, and Brewster's law each appear regularly too. Learn these six and you cover most questions.

Does fringe width change if the setup is put in water?

Yes. Inside a medium of refractive index mu, wavelength becomes lambda/mu, so fringe width becomes beta/mu (fringes get closer). Frequency and colour stay the same; only wavelength and speed change.

What is the difference between fringe width and angular fringe width?

Fringe width beta = lambda*D/d is a length on the screen and depends on D. Angular fringe width theta = lambda/d is an angle and does NOT depend on D. Moving the screen changes beta but not the angular width.

Is a*sin(theta) = n*lambda for bright or dark fringes?

For single-slit diffraction it gives the DARK minima. For double-slit (YDSE) the same form d*sin(theta) = n*lambda gives BRIGHT maxima. Always check single vs double slit first.

What does the 1.22 in the resolving power formula come from?

The 1.22 comes from the diffraction pattern of a circular aperture (the first dark ring of the Airy pattern). It is a fixed constant for round openings like telescope and microscope apertures.