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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
What is the ratio of slit widths when amplitudes of light waves from them have ratio `sqrt(3) : sqrt(2)` ? |
| Answer» `(w_(1))/(w_(2)) = (I_(1))/(I_(2)) = (a^(2))/(b^(2))((sqrt(3))/(sqrt(2)))^(2) = (3)/(2)` | |
| 552. |
Two convex lenses of an astronomical telescope have focal length 1.3 m and 0.05 m respectively. Find the magnifying power and the length of the telescope. |
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Answer» Given: f0 = 1.3 m, fe = 0.05 m To find: i. Magnifying power of telescope (M.P.) ii. Length of telescope (L) Formulae: i. M.P = \(\frac{f_0}{f_e}\) ii. L = f0 + fe Calculation: From formula (i), M.P = \(\frac{1.3}{0.05}\) = 26 From formula (ii), L = 1.3 + 0.05 = 1.35 m |
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| 553. |
Give three basic differences between real image and virtual image. |
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Answer» Real Image : (1) Rays meet actually at the image point. (2) It can be taken on a screen. (3) It is always inverted. Virtual Image : (1) Rays appear to diverge from the image point. (2) It cannot be taken on a screen. (3) It is always erect. |
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| 554. |
A candle flame is held `2 metre` above the water level in a tank `4 metre` deep. If `mu` of water is `4//3`, where will the image of candle flame be seen ? |
| Answer» The image of candle flame is seen due to reflection of light from water surface in the tank. The image will be at a depth of `2 metre` from the surface of water in the tank. | |
| 555. |
Two spectral lines of sodium `D_(1) and D_(2)` have wavelengths of approximetely plane wave on to a slit of width `2 micrometer`. A screen is located `2m` from the slit. Find the spacing between the first maxima of two sodium lines as measured on the screen. |
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Answer» Here, `lambda_(1) = 5890 Å = 5890 xx 10^(-10)m`, `a = 2 mu m = 2 xx 10^(-6)m` `lambda_(2) = 5896 Å = 5896 xx 10^(-10)m, D = 2 m` For the first secondary maxima, `sin theta = (3 lambda_(1))/(2 a) = (x_(1))/(D)` or `x_(1) = (3 lambda_(1) D)/(2 a) and x_(2) = (3 lambda_(2) D)/(2 a)` `:.` Spacing between the first secondary maxima of two sodium lines `= x_(2) - x_(1) = (3 D)/(2 a)(lambda_(2) - lambda_(1))` `= (3 xx 2(5896 - 5890) xx 10^(-10))/(2 xx 2 xx 10^(-6)) = 9 xx 10^(-4)m` |
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| 556. |
A screen is placed `2m` away from the single narrow slit. Calculate the slit width if the first minimum lies `5 mm` on either side of the central maximum. Incident plane waves have a wavelength of `5000 Å`. |
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Answer» Here, distance of the screen from the slit, `D = 2m, a = ?, x = 5mm = 5 xx 10^(-3)m`, `lambda = 5000 Å = 5000 xx 10^(-10)m` For the first secondary minima, ,brgt `sin theta = (lambda)/(a) = (x)/(D)` `:. a = (D lambda)/(x) = (2 xx 5000 xx 10^(-10))/(5 xx 10^(-3)) = 2 xx 10^(-4)m` |
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| 557. |
A laser beam has a wavelength of `7 xx 10^(-7) m` and aperure `10^(-2)m`. The beam is sent to moon at a distance of `4 xx 10^(5) km` from earth. Find the angular spread and areal spread of the beam on reaching the moon. |
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Answer» Here, `lambda = 7 xx 10^(-7) m, a = 10^(-2)m` `D = 4 xx 10^(5) km = 4 xx 10^(8)m` For diffraction at a circular aperture, angular spread `= theta = (1.22 lamda)/(d) = (1.22 xx 7 xx 10^(-7))/(10^(-2))` `= 8.54 xx 10^(-5) radian` Areal spread `= (D theta)^(2)` `= (4 xx 10^(8) xx 8.54 xx 10^(-5))^(2) = 1.197 xx 10^(9)m^(2)` |
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| 558. |
In single slit diffraction pattern, why is intensity of secondary maximum less than the intensity of central maximum ? |
| Answer» The central maximum is due to constructive interference of secondary wavelets from all parts of the slit. With increase in `n` (order of spectrum), the wavelets from lesser and lesser parts of the slit produce constructive interferecne to form secondary maxima. That is why the intensity decreaes. | |
| 559. |
Explain how the intensity of diffraction pattern changes as the order `(n)` of the diffraction bands. |
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Answer» The diffraction pattern due to a single slit consists of a central bright maximum at `O` alingwith alternate secondary minima and maxima on either side. The intensity distribution on the screen is represented in Fig. 6(f).6. The secondary maxima ate the points in between secondary maxima and are of rapidly decreasing intensity. If `I_(0)` is intensity of central bright maximum, then intensity of first secondary maximum, is found to be `I_(0)//22`, and of second secondary maximum is `I_(0)//61` and so on. Note that intensity of central maximum is due to wavelets from all parts of the slit exposed to light. In the first secondary maximum, first two parts of the slit send wavelets in opposite phase, which cancel out. Therefore, intensity of 1st secondary maximum is due to wavelets from only one third part of slit. Similarly, the intensity of second secondary maximum is due to wavelets only from one fifth part of the slit, and so on. |
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| 560. |
Sita has kept a stud consists of diamond. What will she observe? Give reason |
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Answer» The diamond stud appears bright because of total internal reflection. |
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| 561. |
Guna passes a ray light through a glass slab. Which optical phenomenon will take place? What can he observe with reference to wavelength? |
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Answer» When a ray of light enters a glass slab he can observe refraction of light. He observed that wavelength of light decreases. |
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| 562. |
In the minimum deviation position of a prism how are angle of incidence and angle of emergence related ? |
| Answer» Angle of incidence = angle of emergence. | |
| 563. |
A prism is placed in the minimum deviation position. Chari has passed a ray of light at an angle of 45°, then what is the value of angle of emergence? Why? |
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Answer» The angle of emergence = 45°. Since, in the minimum deviation positron, angle of incidence is equal to angle of emergence. |
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| 564. |
Which of the following is not involved in formation of a rainbow? (A) refraction (B) angular dispersion (C) angular deviation (D) total internal reflection |
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Answer» Correct answer is (D) total internal reflection |
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| 565. |
Angles of deviation for extreme colours are given for different prisms. Select the one having maximum dispersive power of its material. (A) 7°, 10° (B) 8°, 11° (C) 12°, 16° (D) 10°, 14° |
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Answer» Correct answer is (A) 7°, 10° |
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| 566. |
magnification -3 indicates.what does it mean ? |
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Answer» A negative sign in the value of the magnification indicates that the image is real. |
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| 567. |
Light with wavelength `535nm` falls normally on a diffraction grating. Find its period if the diffraction angle `35^(@)` corresponds to one of the Fraunhofer maxima and the highest order of spectrum is equal to five. |
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Answer» The diffraction fromula is `d sin theta_(0) = n_(0)lambda` where `theta_(0) = 35^(@)` is the angle of diffraction corresponding to order `n_(0)` (which is not yet known). Thus `d = (n_(0)lambda)/(sin theta_(0)) = n_(0) xx 0.9327mu m` on using `lambda = 0.535mum` For the `n^(th)` order we get `sin theta = (n)/(n_(0)) sin theta_(0) = (n)/(n_(0)) (0.573576)` If `n_(0) -1`, then `n gt n_(0)` is at least `2` and `sin theta gt 1` so `n = 1` is the highest order of diffraction. If `n_(0) = 2` then `n = 3, 4,` but `sin theta gt 1` for `n = 4` thus the highest order of diffraction is `3`. If `n_(0) = 3`, then `n = 4, 5, 6`. For `n = 6, sin theta = 2 xx 0.57 gt 1`, so not allowed while for `n = 5, sin theta = (5)/(3) xx 0.573576 lt 1` is allowed. Thus in this case the highest order of diffraction is five as given. Hence `n_(0) = 3` and `d = 3 xx 0.9327 = 2.7981 ~~ 2.8mu m`. |
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| 568. |
Light with wavelength `lambda = 0.60 mu m` falls normally on a diffraction grating inscribed on a plane surfcae of a plano-curvex cylinderical glass lens with curvature radius `R = 20 cm`. The period of the grating is equal to `d = 6.0 mu m`. Find the distance between the principle maxima of first order located symmetrically in the focal plane of that lens. |
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Answer» For the lens `(1)/(f) = (n-1) ((1)/(R )-(1)/(oo))` or `f =(R )/(n - 1)` For the grating `d sin theta_(1) =lambda` or `sintheta_(1) =(lambda)/(d)` `cosec theta_(1)=(d)/(lambda), cot theta_(1) = sqrt(((d)/(lambda))^(2) - 1)` `tan theta_(1) = (1)/(sqrt(((d)/(lambda))^(2) - 1))` Hence the distance between the two symmetrically placed first order maxima `=2f tan theta_(1) = (2R)/((n -1)sqrt(((d)/(lambda))^(2) - 1))` On putting `R = 20, n = 1.5, d=6.0mu m` `lambda=0.60mu m` we get `8.04 cm`. |
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| 569. |
A plane light wave with wavelength `lambda = 0.57mu m` falls normally on a suface on a surface of a glass `(n = 1.60)` disc which shuts one and a half Fresnel zones for the observation point `P`. What must the minimum thickness of that disc be for the intensity of light at the point `P` to be the highest ? Take into account the interference of light on its passing through the disc. |
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Answer» We follow the argument of `5.103`. We find that the contribution of the first Fresnel zone is `A_(1)=-(4pi i)/(k)a_(0)e^(-ikb)` For the next half zone it is `-(A_(2))/(2)(1 + i)` (The contribution of the remaining part of the `2^(nd)` Fresnel zone will be `-(A_(2))/(2)(1 - i))` If the disc has a thickness `h`, the extra phase difference suffered by the light wave in passing through the disc will be `delta = (2pi)/(lambda) (n - 1)h`. THus the amplitude at `P` will be `E_(P) = (A_(1)-(A_(2))/(2)(1+i))e^(-i delta) -(A_(2))/(2)(1 - i) + A_(3) - A_(4) - A_(5)+...........` `~~((A_(1)(1 -i))/(2))e^(-i delta) + (iA_(1))/(2) = (A_(1))/(2)[(1 - i)e^(-i delta) + i]` The corresponding intensity will be `I = I_(0) (3-2 cos delta - 2 sin delta) = I_(0) (3-2 sqrt(2) sin(delta + (pi)/(4)))` The intensity will be a maximum when `sin(delta + (pi)/(4)) =-1` or `delta + (pi)/(4) = 2kpi + (3pi)/(2)` i.e., `delta = (k + (5)/(8)).2pi` so `h = (lambda)/(n - 1) (k + (5)/(8)), k = 0, 1, 2,.........` Note:- It is not clear why `k = 2` for `h_(min)`. The normal choice will be `k = 0`. If we take `k = 0`, we get `h_(min) = 0.59mum`. |
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| 570. |
Assertion : The sun lookes bigger in size at sunrise and sunset than during day. Reason : The phenomenon of diffraction bends light rays.A. If both, Assertion and Reason are true and the Reason is the correct explaination of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explaination of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - B The sun looks bigger in size at sunrise and sunset because of refraction of light. So the reason, through correct, is not a correct explaination of the assertion. |
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| 571. |
A laser emits a light pulse of duration `tau = 0.13 ms` and energy `E = 10J`. Find the mean pressure extered by such a light pulse when it is focussed into a spot of diameter `d = 10mu m` on a surface perpendicualr to the beam and possessing a reflection coefficient `rho = 0.50`. |
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Answer» The mean pressure `lt p gt` is related to the force fexerted by the beam by `lt p gt xx (pid^(2))/(4) = F` The force `F` equals momentum transferred per second. This is (assuming that photons, not reflected, are absorbed) `2rho(E)/(ctau) +(1-rho)(E)/(ctau) = (1+rho) (E)/(c tau)`. The first term is the momentum transfered on reflection (see problem (261)), the second on absorption. `lt p gt = (4(1_rho)E)/(pid^(2)c tau)` Substituting the values we get `lt p gt = 48.3` atomsphere. |
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| 572. |
In a reference frame `K`a photon of frequency `omega` falls normally on a mirror approaching it with relativistic velocity `V`. Find the momentum inparted to the mirror during the reflection of the photon (a) in the reference frame fixed to the mirroe, (b) in the frame `K`. |
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Answer» (a) In the reference frame fixed to the mirror, the frequency of the photons is, by the Doppler shift formula `overset_(omega) = omega sqrt((1+beta)/(1-beta)) (=omegasqrt(1-beta^(2))/(1-beta))` In this frame momentum imparted to the mirror is `(2cancelh overset_(omega))/(c) = (2cancelh omega)/(c) sqrt((1+beta)/(1-beta))`, (b) In the `K` frame, the incident particle carries a momentum of `cancelh omega//c`and returns with momentum `(cancelh omega)/(c)(1+beta)/(1-beta)` The momentum inparted to the mirror, then, has the magnitude `(cancelh omega)/(c)[(1+beta)/(1-beta)+1] = (2cancelh omega)/(c)(1)/(1-beta)` Hence `beta = (V)/(c)`. |
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| 573. |
If a convex mirror is held in air and then dipped in oil, then what will be the change in its focal length? |
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Answer» Focal length of spherical mirrors are independent of the medium. |
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| 574. |
If a spherical mirror is dipped in water, does its focal length change ? 17. If a thin lens is dipped in water, does its focal length change ? |
| Answer» There would be no change in focal length of the concave mirror held in water as `f` does not depend upon the external meduim in which mirror is held. | |
| 575. |
What is the difference between the virtual images produced by (i) plane mirror (ii) concave mirror and (iii) convex mirror ? |
| Answer» In a plane mirror, virtual image is of same size as the object. In a concave mirror, the virtual image is magnified and in a convex mirror. Virtual image is always diminished in size. | |
| 576. |
What is the advantage of using a parabolic concave mirror over ordinary spherical concave mirror ? |
| Answer» A parabolic concave mirror is free from spherical aberration as well as chromatic aberration. | |
| 577. |
Which property of concave mirror is utilized for using them as shaving mirrors ? |
| Answer» a concave mirror forms a magnified and erect image of the face when it is held close to the face. That is why a concave mirror is prefferred over a plane mirror for shaving. | |
| 578. |
A man standing in front of a special mirror finds his image having a small face, big tummy and legs of normal size. What are the shapes of three parts of the mirror? |
| Answer» Very small head indicates that top part of mirror is convex. A fat body indicates that middle part of the mirror is concave. As legs are of normal size, bottom part of the mirror must be plane. | |
| 579. |
Ray optics is valid, when characteristic dimension ions are (a) much smaller than the wavelength of light (b) much larger than the wavelength of light (c) of the same order as the wavelength of light (d) of the order of one millimetre |
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Answer» (b) much larger than the wavelength of light Ray optics is valid, when characteristic dimensions are much larger than the wavelength of light. |
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| 580. |
A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirrorwillbe(a) 12 feet (b) 3 feet (e) 6 feet (d) any length |
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Answer» (b) 3 feet minimum height of mirror required for seeing full image Height of the man = 3 feet. |
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| 581. |
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to ? How is much a defect of vision corrected ? |
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Answer» This defect is called Astigmatism. It arises because curvature of the plus eye lens refractive system is not the same in different planes. As vertical lines are seen distinctly, the curvature in the vertically plane is enough, but in the horizontal plane, curvature is insufficient. This defect is removed by using a cyclindrical lens with its axis along the vertical. |
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| 582. |
A ray incident at 15° on one refracting surface of a prism of angle 60° , suffers a deviation of 55°. What is the angle of emergance? (a) 95° (b) 45° (c) 30° (d) none of these |
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Answer» (d) none of these A + δ = i + each 60° + 55° = 15° + each e = 115 -15 =100° |
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| 583. |
A ray of light suffers minimum deviation, while passing through a prism of refractive index `1.5` and refracting angle `60^@`. Calculate the angle of deviation and angle in incidence. |
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Answer» Correct Answer - `37.2^(@), 48.6^(@)` `mu = 1.5, A = 60^(@), delta_(m) = ?, i = ?` As `(sin (A + delta_(m))//2)/(sin A//2) = mu` `:. (sin (60^(@) + delta_(m))//2)/(sin 60^(@)//2) = 1.5` `sin (60^(@) + delta_(m))//2 = 1.5 sin 30^(@) = 0.75` `(60^(@) + delta_(m))/(2) = sin^(-1)(0.75) = 48.6^(@)` `delta_(m) = 2 xx 48.6 - 60^(@) = 37.2^(@)` `i = (A + delta_(m))/(2) = (60^(@) + 37.2)/(2) = 48.6^(@)` |
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| 584. |
An object is held in front of a concave mirror between `f` and `C`. The image formed is.A. at `F`B. at `C`C. beyond `C`D. none of the above |
| Answer» Correct Answer - C | |
| 585. |
A beam of light converges to a point P.A lens is placed in the path of the convergent beam `12 cm` from `P`. At what point does the beam converge if the lens is (a) a convex lens of focal length `20 cm`. (b) a concave lens of focal length `16 cm` ? |
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Answer» Here, the point `P` on the right of the lens acts as a virtual object, `:. U = 12 cm, v = ?` As `(1)/(v) - (1)/(u) = (1)/(f) :. (1)/(v) - (1)/(12) = (1)/(20)` `(1)/(v) = (1)/(20) + (1)/(12) = (3 + 5)/(60) = (8)/(60)` `v = 60//8 = 7.5 cm` Image is at `7.5 cm` to the right of the lens where the beam converges. (b) `f = - 16cm, u = 12 cm, :. (1)/(v) = (1)/(f) + (1)/(u) = -(1)/(16) + (1)/(12) = (-3 + 4)/(48) = (1)/(48)` `v = 48 cm` Hence, image is at `48 cm` to the right of the lens, where the beam would converge. |
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| 586. |
An object approaches a convergent lens from the left of the lens with a uniform speed `5 m//s` and stops at the focus. The image.A. moves away from the lens with uniform speed `5 m//s`B. moves away from the lens with uniform accelerationC. moves away from the lens with a non-uniform accelerationD. moves towards from the lens with a non-uniform acceleration |
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Answer» Correct Answer - C When an object approaches a convergent lens from the left of the lens with a uniform speed of `5 m//s`, the image moves away from the lens with a non uniform acceleration. For example. If `f = 20 m` and `v = - 50 m , - 45 m` and `-35 m` , we get `v = 33.3m , 36 m , 40 m` and `46.7m`. Clearly, image moves away from the lens with a non uniform accelaration. Choice ( c) is correct. |
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| 587. |
A double convex lens, made of a material of refractive index μ1, is placed inside two liquids of refractive indices μ2 and μ3, as shown. μ2 > μ1 > μ3. A wide, parallel beam of light is incident on the lens from the left. The lens will give rise to(a) a single convergent beam (b) two different convergent beams (c) two different divergent beams (d) a convergent and a divergent beam |
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Answer» Correct Answer is: (d) a convergent and a divergent beam As μ2 > μ1, the upper half of the lens will become diverging. As μ1 > μ3, the lower half of the lens will become converging. |
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| 588. |
At what thickness will a thick convex-cancave glass lens in air (a) serve as a telescope provided the curvature radius of its convex surface is `DeltaR = 1.5 cm` greater than that of its concave surface? (b) have the optical equal to `-1.0 D` if the curvature radii of its convex and concave surfaces are equal to `10.0` and `7.5 cm` respectively ? |
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Answer» A telescope in normal adjustment is a zero power conbiation of lenses. Thus we require `Phi = O = Ph_(1) + Phi_(2) - (d)/(n) Phi_(1)Phi_(2)` But `Phi_(1) =` Power of the convex surface `= (n - 1)/(R_(0) + DeltaR)` `Phi_(2) =` Power of the concave surface `= -(n - 1)/(R_(0))` Thus, `O = ((n - 1) DeltaR)/(R_(0)(R_(0) + DeltaR)) + (d)/(n) ((n -1)^(2))/(R_(0)(R_(0) + DeltaR))` So `d = (n DeltaR)/(n - 1) = 4.5 cm`. on putting the values. (b) Here, `Phi =- 1 = (.5)/(.1) - (.5)/(.075) + (d)/(1.5) xx (.5 xx .5)/(.1 xx .075)` `=5 - (20)/(3) + (d xx 2)/(3) xx (5 xx 20)/(3) =- (5)/(3) + (200d)/(9)` `= (200d)/(9) = (2)/(3)` or `d =(3//100)m = 3cm`. |
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| 589. |
Answer the following questions : (a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band. (b) In what way is diffraction from each slit related to interference pattern in a double slit experiment ? ( c) When a tiny circular obstacle is placed in the path if light from a distant source, a bright sound seen at the centre of the shadow of the obstacle. Explain why ? ltbr. (d) Two students are separated by a `7 m` partition wall in a room `10 m` high. If both light and sound waves can bend round corners, how is it that the students are unable to see each other even through they can converse easily. (e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures//slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so sommonly used in understanding location and several other properties of images in optical instruments. What is the justification ? |
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Answer» (a) when width (a) of single slit is made double, the half angular width of central maximum which is `lambda//a`, reduces to half. The intensity of central maximum will become `4 times`. This is because area of central diffraction band would become `1//4th`. (b) If width of each slit is of the order of `lambda`, then interference pattern in the double slit experiment is modified by the diffraction pattern from each of the two slits. ( c) This is becasue waves diffracted from the edges of circular obstacle interfere constructively at the centre of the shadow resulting in the formation of a bright spot. (d) For diffraction of waves by obstacle//aperture, through a large angle, the size of obstacle//aperture should be comparable to wavelengt. This follows from `sin theta = lambda//a`. For light, `lambda ~~ 10^(-7)m` and sizze of wall `a ~~ 10m. :. sin theta = (lambda)/(a) = (10^(-7))/(10) = 10^(-8) :. theta rarr 0`. i.e light goes almost unbent. The students are thus unable to see each other. For sound waves of frequency `~~ 1000 Hz, lambda = (v)/(n) = (330)/(1000) = 0.33m` `sin theta = (lambda)/(a) = (0.33)/(10) = 0.033` `:. theta` has a definite values i.e. sound waves bend around the partition. Hence students can cinverse easily. (e) The ray optics assumption is used in understanding loaction and several other properties of image in optical intstruments. This is because typical sizes of paertures invloved in ordinary optical instruments are much larger than the wavelength of light. Therefore, diffraction or bending of waves is of no significance. |
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| 590. |
Calculate the positions of the principle planes and focal points of a thick convex-concave glass lens if the curvature radius of the convex surface is equal to `R_(1) = 10.0 cm` and of the concave surface to `R_(2) = 5.0 cm` and the lens thickness is `d = 3.0 cm`. |
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Answer» The optical power of first convex surface is, `Phi = (P(n - 1))/(R_(1)) = 5D`, as `R_(1) = 10 cm` and the optical power of second concave surface is, `Phi_(2) = ((1 - n))/(R_(2)) =- 10D` So, the optical power of the system, `Phi = Phi_(1) + Phi_(2) -(d)/(n)Phi_(2) =-4D` Now, the distance of the primary principle plane from the vertex surface is given as, `x = ((1)/(Phi))((d)/(n)) Phi_(2)`, here `n_(1) = 1` and `n_(2) = n`. `= (d Phi_(2))/(Phin) = 5cm` and the distance of secondary principle plane from the vertex of second concave surface, `x =-((1)/(Phi))((d)/(n)) Phi_(1) =- (d Phi_(1))/(Phi n) = 2.5 cm` |
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| 591. |
A convex lens of focal legnth `0.2 m` and made of glass `(mu = 1.50)` is immersed in water `(mu = 1.33)`. Find the change in the focal length of the lens. |
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Answer» Correct Answer - `0.58 m` For glass lens in air, `.^(a)mu_(g) = 1.5` and `f_(a) = 0.2 m` `(1)/(f_(a)) = (.^(a)mu_(g) - 1)((1)/(R_(1)) - (1)/(R_(2)))` `(1)/(0.2) = (1.5 -1)((1)/(R_(1)) - (1)/(R_(2)))` `:. (1)/(R_(1)) - (1)/(R_(2)) = 10 cm` For the same lens in water, `(1)/(f_(a)) = ((.^(a)mu_(g))/(.^(a)mu_(w)) - 1)((1)/(R_(1)) - (1)/(R_(2))) = ((1.5)/(1.33) - 1) xx 10` `f_(w) = (1.33)/(1.71) = 0.78 m` Change in focal length `= f_(w) - f_(a) = 0.78 - 0.20` `= 0.58 m` |
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| 592. |
What is the relation between the refractive indices` mu,mu_(1) ` and `mu_(2)` if the behaviour of light rays is shown in Figure. |
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Answer» In Fig., the focal length `f` of the lens is given by `(1)/(f) = ((mu)/(mu_(1) - 1))((1)/(R_(1)) - (1)/(R_(2)))` As the rays pass through the lens without any deviation, therefore, `f = oo` `:. ((mu)/(mu_(1) -1))(1)/(R_(1) - (1)/(R_(2))) = (1)/(oo) = 0` For a double convex lens, `((1)/(R_(1)) - (1)/(R_(2))) is +` `:. (mu)/(mu_(1)) = 1 = 0 or (mu)/(mu_(1)) = 1 or mu = mu_(1)` In Fig.(b), `(1)/(f) = ((mu)/(mu_(2)) - 1)((1)/(R_(1)) - (1)/(R_(2)))` As the lens behave as a diverging lens , therefore `f` is negative As the factor `((1)/(R_(1)) - (1)/(R_(2)))` is positive, `:. ((mu)/(mu_(1)) - 1)` is negative i.e., `((mu)/(mu_(2)) - 1) lt 0 or (mu)/(mu_(2)) lt 1 or mu lt mu_(2)` |
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| 593. |
A short sighted person is wearing specs of power `-3.5 D`. His doctor prescribes a correction of `+ 2.5 D` for his near vision. What is focal length of his distance viewing part and near vision. |
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Answer» Correct Answer - `28.5cm` ; `16.7 cm` For distance viewing part , `P_(1) = - 3.5 D` `:. f_(1) = (100)/(P_(1)) = (100)/(-3.5)cm = - 28.5 cm` For near vision part, `P_(1) + P_(2) = P` `P_(2) = P - P_(1) = 2.5 D - (3.5 D) = 6.0 D` `f_(2) = (100)/(P_(2)) = (100)/(6.0) = 16.7 cm` |
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| 594. |
The far point of a myopic person is `150 cm` in front of the eye. Calculate the focal length and power of a lens required to enable him to see distant objects clearly. |
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Answer» Correct Answer - `-1.5 m, -0.67 D` Here, `v = x = - 150 cm, u = oo, f = ?` From `(1)/(f) = (1)/(v) - (1)/(u) = (1)/(-150) - (1)/(oo)` `f = - 150 cm = - 1.5 m` `P = (1)/(f) = (1)/(-1.5) = -0.67 D` |
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| 595. |
What focal length should the reading spectacles have for a person whose near point is `50cm` ? (NCERT Solved Example) |
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Answer» The book to be read is at `u = - 25 cm` and the image is needed at `v = - 50 cm`. Therefore, focal length `(f)` of the required lens is given by `(1)/(f) = (1)/(v) - (1)/(u) = (1)/(-50) + (1)/(25) = (1)/(50)` `= 50 cm` (convex lens) |
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| 596. |
The near point of a hypermetropic person is `50 cm` from the eye. What is the power of the lens required to enable him to read clearly a book held at `25 cm` from the eye ? |
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Answer» Here, `u = - 25 cm, v = - 50 cm, f = ?` `(1)/(f) = (1)/(v) - (1)/(u) = (1)/(- 50) + (1)/(25) = (1)/(50)` `f = 50 cm, P = (100)/(50) = + 2D` |
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| 597. |
A hypermetropic person whose near point is at `100 cm` wants to read a book at `25 cm`. Find the nature and power of the lens needed. |
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Answer» Here, `u = - 25 cm, v = - 100 cm`, `f = ?` As `(1)/(f) = (-1)/(u) + (1)/(v)` `:. (1)/(f) = (1)/(25) - (1)/(100) = (4 - 1)/(100) = (3)/(100)` `f = (100)/(3) cm = 33.3 cm` `P = (100)/(f) = (100)/(100//3) = 3` dioptre The lens must be converging. |
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| 598. |
An object is placed at a distance of `1.5 m` from a screen and a convex lens is interposed between them. The magnification produced is `4`. What is the focal length of the lens ? |
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Answer» Correct Answer - `0.24 m` Here, the image formed on the screen is real. Therefore, magnification `m = -4`. Let the lens be at a distance `x` from the object. `:. U = - x, v = + (1.5 - x)` As `m = (v)/(u) :. -4 = (1.5 - x)/(- x)` `4x = 1.5 -x, x = 0.3 m` `:. u = - 0.3 m` and `v = 1.5 - 0.3 = 1.2 m` `(1)/(f) = (1)/(v) = (1)/(u) = (1)/(1.2) - (1)/(-0.3) = (5)/(1.2)` `f = (1.2)/(5) = 0.24 m` |
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| 599. |
The focal lengths of the eye piece and objective of a compound microscope are `5 cm and 1 cm` respectively, and the length of the tube is `20 cm`. Calculate magnifying power of microscope when the final image is formed at infinity. The least distance of distinct vision is `25 cm`. |
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Answer» Correct Answer - `70` Here, `f_(e) = 5 cm, f_(0) = 1 cm, L = 20 cm , m = ?` `v_(e) = oo, d = 25 cm` As final image is at infinity, image formed by objective lies at the focus of eye piece. `:. v_(0) = L - f_(e) = 20 - 5 = 15 cm` From `(1)/(u_(0)) = (1)/(v_(0)) - (1)/(f_(0)) = (1)/(15) - (1)/(1) = -(14)/(15)` `u_(0) = -(15)/(14)cm` As final image is at infinity. `:. m = (v_(0))/(u_(0))((d)/(f_(e))) = (15)/(15//14) xx (25)/(5) = 70` |
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| 600. |
The graph in Fig. shows how the inverse of magnification `1//m` produced by a convex thin lens varies with object distance `u`. What was the focal length of the lens used ? .A. `(b)/(c )`B. `/(ca)`C. `(bc)/(a)`D. `( c)/(a)` |
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Answer» Correct Answer - D From `(1)/(v) - (1)/(u) = (1)/(f)` `(u)/(v) - 1 = (u)/(f)` `(1)/(m) - 1 = (u)/(f)` or `(1)/(m) = ((1)/(f))u + 1` which is the equation of a slit whose slope `(1)/(f) = (b)/( c) :. F = ( c)/(b)` |
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