ASIF INSIGHT PHYSICS OPTICS

Name : Roll No. : Topic : Mohammed Asif

Ph : 9391326657, 64606657

Multiple choice questions(only one is correct) 1. From a convex mirror of focal length 6cm, a 3cm tall object is placed at a distance of 7.5cm. Then, a) Image is formed at infinity and it, is point size. b) A virtual, inverted image of size 1.33cm is formed at

10 cm from the mirror. 3

10 cm from the mirror. 3 10 d) A real, inverted image of size 1.33cm is formed at cm from the mirror. 3

c) A virtual erect image of size 1.33cm is formed

2.

PQ is a transparent plate ( µ = 2) having a thickness the light. Then RS is

a) 3.

2 cm

b) 0.62 cm

c) 2 cm

2 cm. RS is the lateral shift in the path of

d) 1.5 cm

In the figure shown what must be the refractive index of the material of the prism so that a ray incident normally on the face AB grazes along the face AC?

a) 2

b)

2

c)

3

d) 1.5

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4.

A glass rod with a hemispherical end has a radius 1cm and its material has refractive index µ=

3 4 . It is immersed in a liquid of refractive index µ = and an object is placed in water 2 3

along the axis of the rod at a distance of 8cm form the curved edge. The image of the object is formed. a) at ∞ b) 2cm towards object c) 2cm away form the object d) at a distance 8cm form object. 5.

A converging beam of ray is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15cm form the lens. If the lens is removed the point of convergence moves 5cm closer to the mounting that holds the lens. The focal length of the lens is a) -30cm b) -15cm c) + 30cm d) + 15cm

6.

Figure shown is a glass paper weight semispherical in shape and of radius 2cm placed in contact with a mark ‘O’ on a white paper. When we view the mark verticlally above AB it will appear at a distance (from AB) of

a) 7.

b) 1.5 cm

c) 1 cm

d) 2 cm

If S1, S2 are pin holes then in the plane of the screen i.e. yz plane the fringe pattern will be

a) 8.

4 cm 3

b)

c)

d)

As shown in figure S1, S2 are two coherent sources emitting light waves of wave length λ . What must be the minimum value of x, so that the point ‘P’ is at maximum intensity?

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a) 4 λ

b) 3 λ

c) 5 λ

d) 2 λ

9.

What must be the minimum thickness of a soap bubble film of µ =1.33 that results in constructive interference in the reflected light, if the film is illuminated by light whose wave O length is 5320 A a) 0.1 µ m b) 1 µ m c) 10 µ m d) 1.5 µ m

10.

If the focal length of the concave mirror is 23cm. What must be the value of x’, so that final image is formed at infinity (given µ of transparent slab is 1.5)

a) 4 cm 11.

b) 15 cm

c) 10 cm

d) 25 cm

In the figure shown is a glass sphere of radius 10cm and made of a transparent material of refractive index 1.5. The final image of an object ‘O’ placed at a distance of 2.5cm is

a) At a distance of 10cm away form B c) At infinity

b) At a distance 100cm form B d) At the position of object

12.

An optical system consists of two convex lenses of focal lengths f1 = 20cm and f2 = 10cm. The distance between the lenses is d = 30cm. An object is placed at a distance of a1 = 30cm form the first lens in the opposite side of the second lens. The distance of image formed form second lens is a) 7.5cm b) 15 cm c) Infinity d) 30cm

13.

A point source S is placed midway between two converging mirrors having equal focal length f as shown in figure. Among the following values of d for which only one image is formed is

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a) 2 f 14.

b) 4 f

c) Both a & b

d) 6 f

A fish which is rising up vertically inside a pond with velocity 4cm/s notices a bird, which is   diving vertically downward. Its velocity appears to the diving bird  µW =  

a) 2cm/s 15.

b) 3cm/s

c) 6cm/s

4 3

d) 9cm/s

A plano convex lens made of a material of refractive index µ has radius of curvature R. If its curved surface is silvered, the focal length of lens is R R R Rµ a) 2 µ b) c) d) 2( µ −1) 2 2

I.

Read the following Passage and answer the questions. Each question in this section has ONLY one correct answer. Fermat’s Principle: Fermat discovered a remarkable principle which we often express in these terms: “A light ray traveling form one point to another will follow a path such that, compared with nearby paths, the time required is either a minimum or a maximum or is stationary (or unchanged) (constant)” To prove the law of refraction form Femat’s principle, consider the diagram, which shows two points A and B in two different media and a ray APB connecting them.

Optical path length = n1 1 + n2  2

= n1 a 2 + x 2 + n2 b 2 + ( d − x ) d ∴ = 0 ⇒ n1 sin i = n2 sin r dx 2

16.

Which is law of refraction If t0 is the time taken for the ray APB in the illustration shown and if C is the velocity of light in free space, then t0 equals

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a)

( n1 1 + n2 2 ) C 1 + 2

b) ( n11 + n2  2 ) C

c)

n11 + n2 2 C

d)

1 n11 + n2 2

17.

If t0 is the time taken for the optical path APB, then, as compared to other near by paths form A to B, t0 the a) Minimum time b) Maximum time c) Same time as any other near by path d) Can be any of the above depending upon the ratio

18.

The word ‘near by paths’ is added in the Fermat’s principle so that you do not take a far-off path such as AP’B as shown in the figure below.

All paths within the hatched region shown can be considered as ‘near by paths’. If t0 is time taken for optical path APB (this is the path with n 1 sin i = n2 sin r), and tS, tQ, tR be respective time value for optical paths through S, Q, R respectively. Then out of tS, tQ, tR a) tQ will be the minimum b) tR will be the minimum c) tS will be the minimum d) None of the above 19.

Suppose we apply Fermat’s principle in reflection, as shown below.

d = 0 ⇒ Gives condition i = r, dx

Let the time taken for such a path APB be t1. Compared to other near-by paths, t1 could be either minimum or maximum or constant. We can work it out of course, but without working it out, we can conclude its behavior on comparing with the case of refraction in the previous question and say whether t1 will show same behavior or opposite behavior to t 0 (of the refracted ray) a) Similar to (Either both minimum or both maximum b) Dissimilar to t0 c) t1 will be constant but t0 is not c) Cannot conclude unless values of dimensions a and b are known. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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20.

When an image is formed at an object by a thin lens, it can be shown that for all rays form object to image through the lens, the time taken is stationary or same. This result is successfully used in a) Measurement of velocity of light b) Using a lens in front of slits in interference experiment c) Construction of compound microscope d) Use of lens as spectacles for correcting myopia

II.

Conditions for permanent interference of light waves coherent sources: To observe the phenomenon of sustained interference of light waves at a place, the following conditions should be fulfilled. (i) The two sources should continuously emit waves of the same length. (ii) For obtaining interference fringes, the amplitudes of the two interfering wave trains should be equal or very nearly equal. (iii) The two sets of interfering wave trains should either have the same phase or a constant difference in phase; that is they must originate form a single parent source of light. (iv) The two sources should be very narrow. (v) The two sources emitting a set of interfering beams should be very close to each other, that is the two interfering wave front must intersect at a very small angle. (vi) For complete annulment or positions of minima, the phase difference between the waves form the two sources should be π or ( 2π +1) and for positions of maxima (reinforcement) is should be 2π or 2nπ .

21.

Refer to condition (i). If the condition is not met; instead the two souces emit only a single pulse each, but do so simultaneously. While we will not be able to observe (with our eyes) the interference pattern on the screen positions, but let us assume that the screen detects the maxima (constructive interference) and minima (destructive interference). Then, a) There will be no maxima or minima b) There will be a central maximum, but no other maxima nor minima c) There will be a central maximum and two minima, but no other. d) There will be various maxima and minima at various positions as per condition no.6

22.

Refer to condition (ii). If the amplitude of one wave-train is half that of the other, then a) The screen will be uniformly bright b) Maxima will be visible but minima will not be visible c) Minima will be visible but maxima will not be visible d) Both maxima and minima will be visible

23.

Refer to condition (iii). Take two cases: case 1: no phase difference (Amplitude of course same). Case 2: There is a phase difference (Amplitude is same as case 1). Then, a) In case 2, the minima will bot be completely dark and also its maxima will be less bright than the maxima in case 1. b) In case 2, the minim will be completely dark but its maxima will be less bright than the maxima in case 1. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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c) In case 2, the minim will not be completely dark but its maxima will have same brightness as case 1. d) The minima of case 2 will be completely dark and its maxima same brightness as case 1. 24.

Refer to condition (iv). This condition is put, as otherwise a) The wave trains may become non-monochromatic. b) The wave trains may have varying phase difference. c) Several overlapping interference patterns will be produced. d) In the interference pattern produced, dark fringes will not be completely dark.

25.

Refer to condition (v). This condition is put as otherwise a) The screen has to be kept very close to the slits in order to observe the interference pattern. b) The distance between adjacent fringes will be very large so that only less number of fringes will be seen. c) The fringe width will be very small, and fringes will not be sharp. d) Central bright fringe cannot be defined

Each question has one or more correct answer. 26. A point object A moves towards a convex mirror with a constant speed v along its optical axis. The speed of the image a) Is always < v b) Is > v if its distance is less than focal length c) Decreases as A goes away from the mirror d) Is may be greater than v for some position of lens. 27.

Large aperture objective is used in telescopes as it helps in a) Increasing field of view b) Decreasing the spherical aberration c) Increasing the resolving power d) Increasing the intensity of image

28.

A cube of side

f is placed with one side along the principal axis of concave mirror of focal 5

length f such that its image will be real, volume will be more than that of object and one face just touches the object then

a) Area of each face of the image is different from that of object b) Area of one face of image is equal to that of object c) Length of the side along the principal axis is

f 4

d) Area of the foace which is farthest form the lens and perpendicular to principal axis is Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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29.

A thin symmetrical double convex lens of poer p is cut into three parts as shown. Then

a) The power of B is

p 2

b) The power of A is

p 2

c) The power of the configuration given in figure(ii) is 0 d) The power of A and C are half of that of B 30.

In a Young’s double slit experiment, let A and B be the two slits. If films of thickness tA and tB and refractive indices µA and µB are place in front of A and B respectively then a) If µA t A = µB t B and t B > t A the central fringe shift towards A b) If µA t A = µB t B and t B > t A the central fringe shifts towards B c) If t A = t B and µA > µB the central fringe shifts toward B d) If t A = t B and µB > µA the central fringe shifts toward B

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KEY 1) 6) 11) 16) 21) 26)

d a b c a b,d

2) 7) 12) 17) 22) 27)

b a a a c a,c,d

3) 8) 13) 18) 23) 28)

b a c b d b,c,d

4) 9) 14) 19) 24) 29)

a a d b b b,c,d

5) 10) 15) 20) 25) 30)

a a a c b a,b,d

Solutions 1.

1 1 1 − = v u f

1  −1  −1 1 −1 −1 − = ⇒ = .  v 7.5  6 v 6 7.5 6 × 7.5 10 v= = cm ( virtual ) 13 .5 3  − 10    v  3  +4 m= = = u − 7.5 9 4 4 10 cm . Size of image = cm, so vitual, erect image of size cm is formed at a distance of 3 3 3

2.

AR = [sec r]t;

AB = cos r AR

R S = sin ( i − r ) AR ⇒ RS =

3.

µ=

sin ( i − r ) t = 0.62 cm cos r

1 1 = = 2 sin C  1     2

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4.

µ 2 µ1 µ 2 − µ1 − = v u R

3 4 3 4      −  2 − 3 = 2 3 ( − 8) v +1 3 1 1 + = ⇒v →∞ 2v 6 6

5.

1 1 1 1 1 1 − = ⇒ − = v u f 15 10 f

F = -30cm

6.

r 2 4 − = µ  3  3 cm from AB.   2

7.

S 2 P − S1 P = n λ is condition for maximum intensity at P. The locus of all such P gives a

hyperbola in 2D motion. So (a) is correct 8.

Condition for minimum x is S1 P − S 2 P = nλ x 2 + 9λ2 − x = λ ⇒ x 2 + 9λ2 = x + λ x + 9λ2 = x 2 + λ2 + 2 x λ 2

8λ2 = 2 x λ ⇒ x = 4λ

9.

For constructive interference (n = 1) 1 λ 5320  2µ t =  n −  λ ⇒ t = = ×10 −10 2 4 µ 4 ×1.33  Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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= 0.1 x 10-6m 10.

In the absence of glass slab the distance of object form concave mirror is x’ + 6 + 15 6

But due to glass slab it is 15 − µ + x ' = F Is the condition for final image to form at infinity 15 −4 + x ' =15 ⇒x ' = 4cm

11.

Lens refraction formula at the nearest surface 1.5 1 0.5 1 − = = v − 2.5 +10 20 1.5 −10 1 1.5 − 200 + 25 −175 = + ⇒ = = v 25 20 v 500 500 500 ×1.5 v= = −4.29 175

At second surface

u = −24 .29 ; v ' =? 1 1 .5 −0.5 1 + = = v ' 24 .29 −10 20 1 1 = −0.06 +0.05 ⇒ = −0.01 v' v' 1 1 = = −100 cm v' 0.01

12.

(

(

13.

)

1 1 1 st − = 1 lens v u F1 1 1 1 nd − = 2 lens v u F2

⇒v = 60 cm

)

⇒ v = 7.5 cm

d = 2f

d = 4f

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Not possible in case d = 6f 14.

15.

v F 1 g + µ v B1 g = v B , F 4 4 + µ v B1 g = 16 ⇒ ×12 ⇒ x = 9 cm 3 R 2µ

16.

Dimensionally other are wrong. Also,     t 0 = t1 + t 2 = 1 + 2 = 1 + 2 v1 v 2  C   C       n1   n2  n  + n2 2 = 1 1 C

17.

If you work out

d 2 = dx 2

(a

n1a 2 2

+x

d 2 d x2

+

) (b

3 2 2

⇒ Always positive.

∴x given

n2 b 2 2

+d−x

)

−2

3 2

d =0 dx

18.

Is the value at which t is minimum. ∴Optical path  = n11 + n2l2 is minimum ⇒ t0 = minimum As proved in the previous question to is minimum.

19.

Take the refraction case. Treat n1 = n2. Geometrically both cases are same.

20.

So it does not introduce any phase difference between converging rays.

21.

Since only a single pulse each, a path difference of

λ 2

alone will be registered, leading to one

minimum on either side of central maximum. 22.

The only disadvantage will be minima will not be completely dark, but lest bright than the rest and maxima of course will be brightest than the rest.

23.

The phase difference causes only a shift in the fringe pattern’s position on screen.

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24.

A broad source is equivalent to a large number of narrow sources lying side by side. Each set of these sources will produce an interference pattern of its own, which will overlap one another so that a fringe system is lost and general illumination results.

25.

β=

λD d

, if d is large

β is small ⇒ interference bands will be very close to each other, will not be sharp, may even overlap.

26.

Ans: (b) and (d)

27.

Ans: (a), (c), (d)

28.

Ans: (b), (c), (d) For the image to touch the object arc face must be at a distance of 2f For the other face u = − 1

1

9 f 5

1

So v + u + f v=

9 f 4

9f f f −2f = 4 4 v 5 m =− =− u 4 f 5 f So side = − = 5 4 4 2 f So area = 16

So length =

29. 30.

Obvious, in case of (c ) one is converging lens and other is diverging lens. Path difference ∆ x = ( µ A − 1) t A − ( µ B − 1) t B = ( µ At A − µ B t B ) + ( t B − t A )

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