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When a ray of light enters a glass slab from air,
NOTE : When the ray enters a glass slab from air, its frequency remains unchanged.
Since glass slab in an optically denser medium, the velocity of light decreases and therefore we can conclude that the wavelength decreases.
(∵ v = vλ)
A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach on the face BC if
The Phenomenon of total internal reflection takes place during reflection at P.
∴ sin θ should be greater than 8/9.
In Young’s double-slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is
A ray of light from a denser medium strike a rarer medium at an angle of incidence i (see Fig). The reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r’ The critical angle is
Applying Snell's law at P, we get
From (i) and (ii)
C = sin –1 (tan r)
Two coherent monochromatic light beams of intensities I and 4 I are superposed. The maximum and minimum possible intensities in the resulting beam are
Spherical aberration in a thin lens can be reduced by
Spherical aberration occurs due to the inability of a lens to converge marginal rays of the same wavelength to the focus as it converges the paraxial rays. This can be done by using a circular annular mask over the lens.
A beam of light of wave length 600 nm from a distance source falls on a single slit 1 mm wide and a resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of central bright fringe is
The distance between the first dark fringe on either side of the central maximum = width of central maximum
An isosceles prism of angle 120° has a refractive index 1.44. Two parallel monochromatic rays enter the prism parallel to each other in air as shown. The rays emerge from the opposite faces
Applying Snell's law at P,
∴ δ = r – 30° = sin – 1 ( 0.72) – 30°
∴ The rays make an angle of
2δ = 2 [sin – 1(0.72) – 30°] with each other..
A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing
NOTE : A convex mirror and a concave lens always produce virtual image.
Therefore, option (b) and (d) are not correct. The image by a convex lens is diminished when the object is placed beyond 2f.
Let u = 2f + x
But u + v = 1 (given)
⇒ (2f + x)2 < f + x. The above is true for f < 0.25 m.
The focal lengths of the objective and the eye piece of a compound microscope are 2.0 cm and 3.0 cm, respectively.
The distance between the objective and the eye piece is 15.0 cm. The final image formed by the eye piece is at infinity.
The two lenses are thin. The distance in cm of the object and the image produced by the objective, measured from the objective lens, are respectively
Here fo = 2 cm and fe = 3 cm.
Using lens formula for eye piece
But the distance between objective and eye piece is 15 cm (given)
∴ Distance of image formed by the objective = v = 15 – 3 = 12 cm.
Let u be the object distance from objective, then for objective lens
Consider Fraunhoffer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum the phase difference (in radians) between the wavelets from the opposite edges of the slit is
Path difference between the opposite edges is λ.
For a phase difference of 2π we get a path diff. of λ.
In an interference arrangement similar to Young’s doubleslit experiment, the slits S1 and S2 are illuminated with coherent microwave sources, each of frequency 106 Hz. The sources are synchronized to have zero phase difference.
The slits are separated by a distance d = 150.0 m. The intensity I (θ) is measured as a function of θ, where θ is defined as shown. If I0 is the maximum intensity, then I (θ) for 0 < θ < 90° is given by
We know that
For θ = 90° ; I(θ) = Io cos2 (∞)
For θ = 0°
I (θ) = I0
I (θ) is not constant.
Alternatively, when θ is zero the path difference between wave originating from S1 and that from S2 will be zero. This corresponds to a maxima.
A concave lens of glass, refractive index 1.5 h as both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a
For concave lens as shown in figure in this case
R1 = – R and R2 = + R
⇒ f = + 3.5 R
NOTE : The positive sign shows that the lens behaves as convergent lens.
Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm. If yellow light is replaced by X– rays, then the observed pattern will reveal,
For diffraction pattern to be observed, the dimension of slit should be comparable to the wave length of rays. The wavelength of X-rays (1 – 100 Å) is less than 0.6 mm.
A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate as shown in Figure.
The observed interference fringes from this combination shall be
Locus of equal path difference are lines running parallel to axis of the cylinder. Hence straight fringes will be observed.
A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids L1 or L2 having refractive indices μ1 and μ2 respectively (μ2 > μ1 > 1). The lens will diverge a parallel beam of light if it is filled with
If μ2 > μ1, the concave lens maintains its nature otherwise the nature of the lens will be reversed.
So, the lens should be filled with L2 and immerse in L1.
A point source of light B is placed at a distance L in front of the centre of a mirror of width 'd' hung vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown in fig. The greatest distance over which he can see the image of the light source in the mirror is
From the ray diagram.
In ΔANM and ΔADB
∠ADB = ∠ANM = 90°
∠MAN = ∠BAN (laws of reflection)
Also ∠BAN = ∠ABD ⇒ ∠MAN = ∠ABD
∴ ΔANM is similar to ΔADB
So, required distance = d + d + d = 3d.
A diverging beam of light from a point source S having divergence angle α , falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and the refractive index n, then the divergence angle of the emergent beam is
The incident and emergent ray of a glass slab are parallel therefore, the angle remains the same.
A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2(n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence αmax such that the ray comes out only from the other surface CD is given by
See figure. The ray will come out from CD if it suffers total internal reflection at surface AD, i.e., it strikes the surface AD at critical angle C ( the limiting case).
Applying Snell's law at P
Applying Snell's law at Q
n2 sin α = n1 cos C
In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern
When slits are of equal width
When one slit's width is twice that of other
In a compound microscope, the intermediate image is
NOT E : The intermediate image in compound microscope is real, inverted and magnified.
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between the resultant intensities at A and B is
Applying eq. (1) when phase difference is π/2
Again applying eq. (1) when d phase difference is π
In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by
A ray of light passes through four transparent media with refractive indices μ1, μ2, μ3 and μ4 as shown in the figure.
The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have
Applying Snell's law at P,
Applying Snell's law at Q,
Again applying Snell's law at R
Multiplying (i), (ii) and (iii), we get
µ4 = µ1
NOTE : If the emergent ray is parallel to incident ray after travelling a number of parallel interfaces then the refractive index of the first and the last medium is always same.
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer
There will be no refraction from P to Q and then from Q to R (all being identical). Hence the ray will now have the same deviation.
An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is
For the image of point P to be seen by the observer, it should be formed at point Q.
NS = QS = 2h
∴∠NQS = 45°
∴ r = 45°
Now in ΔQMA,
∠MQA = 45°
∴ MA = QA = h
In Δ PMB,
PM2 = 4h2 + h2 = 5h2
From (i) and (ii)
Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams.
Since both surfaces have same radius of curvature on the same side, no dispersion will occur.
In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wave-lengh t λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is
Path difference = (µ – 1) t = n λ;
For minimum t, n = 1; ∴ t = 2 λ
Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle 30° at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is
Maximum number of reflection
where x = 0.2 tan 30° = 0.2/√3.
In the adjacent diagram, CP represents a wavefront and AO & BP, the corresponding two rays. Find the condition on θ for constructive interference at P between the ray BP and reflected ray OP.
Path difference between the two rays reaching P is
For constructive interference, path difference should be nλ
The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image.
Convex lens forms the image at I1. I1 is at the second focus of convex lens. Size of I1 = 2 cm.
I1 acts as virtual object for concave lens. Concave lens forms the image of I1 and I2.
For concave lens,
or v = 5 cm = Distance of I2 from concave lens.
or size of image due to concave lens = 2.5 cm
A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of μg would be
A beam of white light is incident on glass air interface from glass to air such that green light just suffers total internal reflection. The colors of the light which will come out to air are
∴sin C ∝ λ
For higher value of λ, the angle C also increases
An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation
NOTE : For minimum deviation, incident angle is equal to emerging angle and QR is parallel to base.
Monochromatic light of wavelength 400 nm and 560 nm are incident simultaneously and normally on double slits apparatus whose slits separation is 0.1 mm and screen distance is 1m. Distance between areas of total darkness will be
At the area of total darkness minima will occur for both the wavelengths.
by inspection for m = 2, n = 3 and for m = 7, n = 10, the distance between them will be the distance between such points.
put n2 = 10, n1 = 3
On solving we get, Δs = 28mm.
A source emits sound of frequency 600 Hz inside water. The frequency heard in air will be equal to (velocity of sound in water = 1500 m/s, velocity of sound in air = 300 m/s)
NOTE : Frequency does not change with change of medium.
A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of virtual image from the surface is
The rays coming from the point object fall on the glassair interface normally and hence pass undeviated.
Therefore if we retrace the path of the refracted rays backwards, the image will be formed at the centre only.
In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is
A convex lens is in contact with concave lens. The magnitude of the ratio of their focal length is 2/3. Their equivalent focal length is 30 cm. What are their individual focal lengths?
f1: focal length of convex lens.
f1 = 10 cm, f2 = –15 cm
A container is filled with water (μ = 1.33) upto a height of 33.25 cm. A concave mirror is placed 15 cm above the water level and the image of an object placed at the bottom is formed 25 cm below the water level. Focal length of the mirror is
The image I ' for first refraction (i.e., when the ray comes out of liquid) is at a depth of
Now, reflection will occur at concave mirror. For this I' behaves as an object.
∴ u = – (15 + 25) = – 40 cm
Where 25/1.33 is the real depth of the image.
Using mirror formula we get
Focal length of the plano-convex lens is 15 cm. A small object is placed at A as shown in the figure. The plane surface is silvered. The image will form at
The focal length f of the equivalent mirror is
Since f has a positive value, the combination behaves as a converging mirror.
According to mirror formula
Negative sign indicates that the image is 12 cm in front of mirror.
The graph shows relationship between object distance and image distance for a equiconvex lens. Then, focal length of the lens is
We know that in case of a convex lens when object is placed at C ', the image is obtained at C. This situation is represented in the graph by the point corresponding to u = –10 cm, v = 10 cm.
Therefore R = 10 cm ⇒ R/2 = 5 cm = f
Lens formula is
(for maximum error in f)
Therefore, the focal length = (5.00 ± 0.05) cm.
Rays of light from Sun falls on a biconvex lens of focal length f and the circular image of Sun of radius r is formed on the focal plane of the lens. Then
In an experiment to determine the focal length (f) of a concave mirror by the u - v method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then,
As shown in the figure, when the object (A) is placed between F and C, the image (I) is formed beyond C. It is in this condition that when the student shifts his eyes towards left, the image appears to the right of the object pin. (Image distance > object distance)
A ray of light traveling in water is incident on its surface open to air. The angle of incidence is θ, which is less than the critical angle. Then there will be
The ray is partly reflected and partly refracted ∠MOB = 180 – 2θ
But the angle between refracted and reflected rays is ∠POB. Clearly is ∠POB is less than ∠MOB.
Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be
For minimum deviation the ray in the prism is parallel to the base of the prism. This condition does not depend on the colour (or wave length) of incident radiation. So in both the cases, by geometry, r = 30º. So (a) is correct option.
A light beam is travelling from Region I to IV (figure). The refractive index in regionals I, II, III and IV are and respectively. The angle of incidence q for whichthe beam just misses entering region IV is –
For refraction at parallel interfaces
A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4.3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m/s2.]
Consider the activitiy A to B
Applying v2 - u2 = 2as
The velocity of ball as perceived by fish is
A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is
Focal length of the biconvex lens is 15 cm. A small object is placed at a distance of 30cm from the lens i.e. at a distance of 2f. Therefore the image should form at 30cm from the lens at I1.
But since the ray strike the plane mirror before reaching I1, the image I1 acts as the virtual object for reflection on plane mirror kept at a distance of 20 cm from it.
It should produce an image I2 but as the ray encounters the lens, it gets refracted and the final image is formed at I3. For the last refraction from the biconvex lens, u = 10 cm.
Applying lens formula
Therefore a real image is formed at a distance of 16 cm from the plane mirror.
A light ray travelling in glass medium is incident on glassair interface at an angle of incidence θ. The reflected (R) and transmitted (T) intensities, both as function of θ, are plotted. The correct sketch is
When the light is incident on glass - an interface at an angle less than critical angle a small part of light will be reflected and most part will be transmitted.
When the light is incident greater than the critical angle, it gets completed reflected (total internal reflection) These characteristics are depicted in option (c).
A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surface are of the same radius of curvature R = 14 cm. For this biconvex lens, for an object distance of 40 cm, the image distance will be
The focal length (f1) of the lens with n = 1.5 is given by
The focal length (f2) of the lens with n = 1.2 is given by
The focal length F of the combination is
Applying lens formula for the combination of lens
⇒ V = 40 cm
Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are bG, bR and bB, respectively. Then,
A ray of light travelling in the direction is incident on a plane mirror. After reflection, it travels along the direction The angle of incidence is
∴ 180° – 2a = 120°
∴ a = 30°
option (a) is correct
In the Young’s double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is
The intensity I is given as
where Io is the peak intensity
For a phase difference of 2π the path difference is λ
∴ For a phase difference of the path difference
option (b) is correct.
A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is
Applying Snell’s law at point P
n1 sin θ = n2 sin 90°
A parallel beam of light is incident from air at an angle a on the side PQ of a right angled triangular prism of refractive index n = √2. Light undergoes total internal reflection in the prism at the face PR when α has a minimum value of 45°.
The angle θ of the prism is
Applying Snell's law at A
Applying Snell's law at B
√2 sin C = 1 × sin 90°
∴ C = 45° … (ii)
In Δ AMB, 90° + θ + r1 + (90° – C) = 180° (From fig.)
∴ θ = 15°
A small object is placed 50 cm to the left of a thi n con vex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm.The mirror is tilted such that the axis of the mirror is at an angle θ = 30° to the axis of the lens, as shown in the figure.
If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are
Therefore object distance for mirror is 25 cm and object is virtual
∴ v = – 50 cm The image I would have formed as shown had the mirror been straight. But here the mirror is tilted by 30°.
Therefore the image will be tilted by 60° and will be formed at A.
Here MA = 50 cos 60° = 25 cm
and I¢A = 50 sin 60° = 25√3 cm