Class 12 Physics, Chapter 9 (NCERT). 15 questions by concept, 2 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
A plane mirror turns through 2° while an incident ray remains fixed. By how much does the reflected ray turn in the plane of incidence?
(c) 4 degrees
The normal rotates with the mirror. The incident and reflected angle changes together give twice the mirror rotation.
An astronomical telescope uses two converging lenses of focal lengths 144 cm and 6 cm. What is their separation when the final image is at infinity?
(d) 150 cm
The objective second focus and eyepiece first focus coincide. Their distances from the two lenses add: S=fo+fe.
At an air-glass spherical interface, the correct equation is (3/2)/v+1/100=1/40, with all distances in cm. A student drops the factor 3/2 and gets v=200/3 cm. What is the corrected v?
(d) 100 cm
Subtract 1/100 from 1/40 to obtain 3/200. Then (3/2)/v=3/200 gives v=100 cm.
For a concave mirror with f=-20 cm and u=-30 cm, a student writes 1/v-1/u=1/f and obtains v=-12 cm. Which replacement is correct?
(a) 1/v+1/u=1/f, giving v=-60 cm
Reflection uses 1/v+1/u=1/f. Hence 1/v=-1/20+1/30=-1/60 cm^-1.
The diagram gives an oblique incident ray and a parallel-sided glass slab surrounded by air. Which relation holds between the final emergent ray and the incident ray?
(c) They are parallel, with a lateral displacement
At the parallel exit face, Snell law reverses the first-face angular change because the external medium is again air. The emerging ray is parallel but shifted.
Light of vacuum wavelength 600 nm and frequency ν enters a stationary transparent medium of refractive index 3. What are its frequency and wavelength in the medium?
(d) ν and 200 nm
Frequency remains ν. Speed becomes c/n, so λ=(c/n)/ν=600/n nm.
Which core/cladding index ordering can support total internal reflection of suitable rays travelling inside an optical fibre?
(a) Core 1.50, cladding 1.40
A core ray must meet a lower-index cladding at incidence above the critical angle. Reversing the indices prevents that denser-to-rarer transition.
A student assigns R1=R2=+20 cm to a double convex lens of index 1.50 in air and concludes its power is zero. How should this be corrected?
(b) Set R2=-20 cm; the focal length is +20 cm
The surface centres lie on opposite sides. Their signed reciprocal difference is 1/20-1/(-20)=1/10 cm^-1; multiplying by 0.5 gives f=20 cm.
A concave lens has f=-21 cm and a real object at u=-14 cm. A student uses 1/v+1/u=1/f and predicts v=+42 cm. Which correction is valid?
(d) Use 1/v-1/u=1/f; v=-8.4 cm
For transmission, 1/v=1/f+1/u=-1/21-1/14=-5/42 cm^-1. Thus v=-8.4 cm, a virtual image.
Thin lenses with focal lengths +20 cm and -15 cm are placed in contact, on the same axis, in air. What is their signed equivalent focal length?
(b) -60 cm
Add signed powers: 1/F=1/p-1/n=(n-p)/(pn). Thus F=pn/(n-p); the sign determines convergence or divergence.
Light travels left to right through lenses A and B. Lens A alone would form an image 15 cm to its right; B is 20 cm to the right of A. What signed object distance u_B must be used at B?
(a) -5 cm
Translate the reference origin from A to B: u_B=v_A-separation=15-s. A positive value is a virtual object for B, a negative value a real object.
White light traverses an ordinary glass prism in air in the visible normal-dispersion regime. Which comparison is correct?
(a) Violet deviates more than red because its refractive index is greater
Normal dispersion gives n_violet>n_red. Greater relative index gives greater prism deviation under corresponding conditions.
A student claims a prism always gives minimum deviation when the ray enters its first face normally. Which correction is valid?
(a) Minimum deviation requires symmetric passage: i=e and r1=r2=A/2
Both interfaces contribute. Ray reversibility and Snell law give the symmetric minimum, generally at nonzero incidence.
A compound microscope objective gives linear magnification magnitude 5 and its eyepiece gives angular magnification 6. What is the total magnification magnitude?
(c) 30 times
The objective first enlarges the object into an intermediate image; the eyepiece magnifies that image angularly. The two factors multiply.
Half of an ideal concave mirror aperture is covered. Rays from every object point still reach the uncovered part. Compared with the original real image, what is observed?
(d) The whole image remains at the same position, with lower brightness
Light from each object point reaches many parts of the mirror. The uncovered part still receives rays from every object point, as the stem states.
For an ideal mirror, those remaining rays still converge at the same image positions. Covering the aperture does not change the curvature, focal length or object distance.
The whole real image remains at the same position and size, but fewer reflected rays make it less bright. Mirror halves do not correspond to image halves.
Wrong
A screen behind a plane mirror can catch the image I see there.
Why it feels right. The reflected image appears to occupy a position behind the mirror.
Right
A screen can capture an image where light rays actually converge. A plane mirror produces a virtual image.
Worked example
