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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.
| 51. |
An object 1 cm high is held near a concave mirror of magnification 10. How tall will be the image? |
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Answer» Correct Answer - `+-10 cm` Here, `h_(1) = 1 cm, h_(2) = ? m= +-10` From `m = (h_(2))/(h_(1)), +-10 = (h_(2))/(1cm) , h_(2) = +- 10 cm`. |
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| 52. |
A concave mirror produces 10 cm long image of an object of height 2 cm. What is the magnification produced? |
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Answer» Correct Answer - `+-5` `m = +-(h_(2))/(h_(1)) = +-(10cm)/(2cm) = +-5` |
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| 53. |
Which of the following lenses would you prefer to use while reading small letters found in a dictionary ?A. Convex lens of focal length 50 cmB. A concave lens of focal length 50 cmC. A convex lens of focal length 5 cmD. A concave lens of focal length 5 cm. |
| Answer» For reading small letters in a dictionary, we need to use a convex lens of smaller focal length. Choice ( c) is correct. | |
| 54. |
State the expression for linear magnification of a concave mirror in terms of object distance and image distance. |
| Answer» `m=(h_(2))/(h_(1))= -(v)/(u)= - ("image distance")/("object distance")` | |
| 55. |
How are power and focal length of a lens related ? You are provided with two lenses of focal length `20 cm` and `40 cm` respectively. Which lens will you use to obtain more convergent light ? |
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Answer» `P = (1)/(f)`. For obtaining more convergent light, power of lens must be larger. Therefore, its focal length must be smaller, Therefore, the lens of `f = 20 cm` will provide more convergent light. |
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| 56. |
Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself ? Explain using a diagram. |
| Answer» When a rectangular glass slab is immersed in any medium, the ray incident on the slab suffers two refractions, and emerges parallel to itself but displaced//shifted laterally as shown Fig. This is because deviation suffered by the ray in first refraction at one face of slab is equal and opposite to deviation suffered by the ray in 2nd refraction at the other face of the slab. | |
| 57. |
The mirror formula is `(1)/(f) = (1)/(v) + (1)/(u)` where symbols have standard meaning. How will focal length of mirror change when u is changed by moving the object towards or away from the mirror? |
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Answer» Focal length (f) of a mirror does not depend upon u or v. It cannot be changed by moving the object towards or away from the mirror. This is because, when we change object distance (u), the image distance (v) changes, keeping (f) constant. Infact, focal lenght of mirror is half the radius of curvatur (R) of the mirror, i.e., `f = R//2`. Focal length will change with change in radius of curvature of the mirror. |
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| 58. |
What is the difference between lens formula and mirror formula? |
| Answer» The lens formula is `(1)/(f)=(1)/(v)-1/u` and the mirror formula is `(1)/(f)=(1)/(v)+1/u` | |
| 59. |
The formula for linear magnification of a spherical mirror is `m=(h_(2))/(h_(1))= (-v)/(u)`. What determines the sign of m? What is the signification of the sign? |
| Answer» In the formula, `m=(h_(2))/(h_(1))=(-v)/(u)`, the sign of m is determined by the signs of `h_(1)` and `h_(2)`. When m is positive, the image is virtual and erect. When m is negative, the image is real and inverted. | |
| 60. |
If glass slab is replaced by a hollow slab filled with water, will the net deviation change? For given angle of incidence, will the deviation of the ray in first refraction increase `3s` or decreases compared to the one in glass slab? |
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Answer» When glass slab is replaced by a hollow slab filled with water, the net deviation of the ray continues to be zero. The deviation of the ray in first refraction in slab filled with water will become less than deviation of the ray in first refraction in glass slab. This is because `mu = sin i/sin r`, and `mu_w lt mu_g` therefore, for given `angle i, sin r ` and hence r will be more in water slab. The deviation (i-r) in water slab will less than that in glass slab. |
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| 61. |
In the mirror formula, `(1)/(f)=(1)/(v)+1/u,f` does not change when u is changed. |
| Answer» The mirror formula is `(1)/(v) + 1/u =(1)/(f)`, where the symbols have standard meaning. When u is changed, v changes, but f remains constant. This is because focal length of mirror depends only on radius of curvature of the mirror. | |
| 62. |
A student very cautiously traces the path of a ray through a glass slab for different values of the angle of incidence `(/_i)`. He then measures the corresponding values of the angle of refraction `(/_r)` and the angle of emergence `(/_e)` for every value of the angle of incidence. On analysing these measurements of angles, his conclusion would beA. `/_i gt /_r gt /_e`B. `/_i= /_e gt /_r`C. `/_i lt /_r lt /_e`D. `/_i= /_e lt /_r` |
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Answer» (b) The correct path of a ray through a glass slab is shown in Fig. As net deviation through the glass slab is zero, therefor `/_i = /_i`. At `B`, the ray goes from a rarer to a denser medium, bends towards normal. Therefore `/_r lt /_i`. or `/_i gt /_r` Hence we conclude that `/_i=/_e gt /_r`. |
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| 63. |
You are given a concave mirror of focal length 30 cm. How can you form a real image of the size of the object using this mirror? |
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Answer» For imgae of size of object `|u|=|v|=2 f= 2xx 30 =60 cm` The object must be held at a distance of `60 cm` from the concave mirror. |
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| 64. |
A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement ? If yes, where shall the object be placed in each case for obtaining these images ? |
| Answer» Yes, a convex lens of focal length `20 cm` can produce a magnified virtual as well as real image. The statement is correct. For magnified virtual image, the object must be held at a distance less than `20 cm` from the lens. For magnified real image, the object must be held at a distance between 20 cm and 40 cm from the lens. | |
| 65. |
Light enters from air into diamond, which has a refractive index of `2.42`. Calculate the speed of light in diamond. The speed of light in air is `3 xx10^(8)m//s`. |
| Answer» Here, `n=2.42,v=?,c= 3 xx 10^(8) m//s` As `n=c/v, v=c/n=(3xx10^(8))/(2.42)=1.24 xx 10^(8) m//s`. | |
| 66. |
The refractive index of diamond is `2.47` and that of glass is `1.51`. How much faster does light travel in glass than in diamond ? |
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Answer» Correct Answer - 1.635 time `(v_(g))/(v_(d)) = (n_(d))/(n_(g)) = (2.47)/(1.51) = 1.635`, Hence light travels in diamond 1.635 times faster than in glass. |
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| 67. |
A coin in a glass beaker appears to rise as the beaker is slowly filled with water. Why? |
| Answer» It happens on account of refraction of light. A ray of light starting from the coin goes water to air and bends away from normal. Therefore, bottom of the beaker on which the coin lies appears to be raised | |
| 68. |
Calculate speed of light in water of refractive index `4//3`. Given speed of light in air `= 3 xx 10^(8) m//s`. |
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Answer» Correct Answer - `2.25 x 10^(8)m//s` Here, `.^(a)n_(w) = 4//3, v_(a) = 3 xx 10^(8) m//s, v_(w) = ?` From `.^(a)n_(w) = (v_(a))/(v_(w))` `v_(w) = v_(a)/(.^(a)n_(w)) = (3xx 10^(8))/(4//3) = 2.25 xx 10^(8) m//s` |
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| 69. |
(a) What is the speed of light in water of refractive index `4//3`? (b) Light travels in a medium with a velocity of `2xx10^(8)m//s`. What is refractive index of the medium? |
| Answer» (a) From `v=c/n, v=(3xx10^(8)m//s)/(4//3)=2.25xx10^(8)m//s`. , (b) From `n=c/v=(3xx10^(8)m//s)/(2xx10^(8)m//s)=1.5`. | |
| 70. |
Figure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is A. `sqrt(3)//sqrt(2)`B. `sqrt(2)//sqrt(3)`C. `1//sqrt(2)`D. `sqrt(2)` |
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Answer» Correct Answer - A By definition, refractive index of medium B w.r.t. A is `n = (sin60^(@))/(sin45^(@)) = (sqrt(3)//2)/(1//sqrt(2)) = (sqrt(3))/(sqrt(2))`. |
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| 71. |
Can light travel in vacuum ? If yes, with what speed ? |
| Answer» Yes, `3 xx 10^(8) m//s`. | |
| 72. |
A ray of light is travelling from glass to air. The angle of incidence in glass is `35^(@)`, and angle of refraction in air `60^(@)`. What is the refractive index of glass w.r.t. air ? |
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Answer» Correct Answer - 1.51 Here, `i= 35^(@), r = 60^(@), .^(a)n_(g) = ?` From `.^(a)n_(g) = (sini)/(sinr) =(sin60^(@))/(sin35^(@)) = (0.866)/(0.5736) = 1.51` |
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| 73. |
What is the basic difference between reflection and refraction of light? |
| Answer» Reflaction is the phenomenon of change in the path of light without any change in medium. Refraction is the phenomenon of change in the path of light in going from one medium to another. | |
| 74. |
How much time will light take to cross 2 mm thick glass pane if refractive index of glassis `3//2`? |
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Answer» Here, `t = ?, x = 2 mm =2xx10^(-3)m, n = 3/2` From `n = c/v, v = c/n = (3xx10^(8))/(3//2) = 2xx10^(8) m//s` `t = x/v = (2xx10^(-3))/(2xx10^(8)) = 10^(-11)s` |
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| 75. |
A ray of light passes from glass to air at an angle of `19.5^@`. Calculate the angle of refraction, given refractive index of glass w.r.t. air is `3//2`. |
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Answer» As the ray passes from glass to air, `:. n = sin r/sin i=1.5` `sin r = n sin i=1.5 sin 19.5^@` `=3/2 xx 0.3333=0*5` `r =30^@`. |
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| 76. |
A pond of depth 20 cm is filled with water of refractive index `4//3`. Calculate apparent depth of the tank when viewed normally. |
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Answer» Here, real depth of pond, `x = 20 cm`, refractive index of water, `.^(a)n_(w) = 4/3` apparent depth, `y = ?` As `.^(a)n_(w) = x/y`, `y = x/(.^(a)n_(w)) = 20/(4//3) = 20xx(3)/(4) = 15 cm` |
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| 77. |
Light travels through water with a speed of `2.25 xx 10^(8)m//s`. What is the refractive index of water ? Given speed of light in vacuum `= 3xx10^(8)m//s`. |
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Answer» Here, speed of light in water, `v = 2.25 xx 10^(8)m//s` speed of light in vacuum, `c = 3 xx 10^(8)m//s`, refractive index, `n = ?` From `n = c/v`, `n = (3xx10^(8))/(2.25xx10^(8)) = 1.33` |
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| 78. |
With respect to air, the refractive index of ice is `1*31` and that of rock salt is `1*54`. Calcualte the refractive index of rock salt w.r.t ice. |
| Answer» Here, `.^(a)n_(i)=1.31` and `.^(a)n_(r) =1.54 , .^(i)n_(r)=?` Now, `.^(i)n_(r) =(.^(a)n_(r))/(.^(a)n_(i))=(1.54)/(1.31)=1.17`. | |
| 79. |
The correct sequencing of angle of incidence, angle of emergence, angle of refraction and lateral displacment shown in the following diagram by digits 1,2,3 and 4 is : A. 2,4,1,3B. 2,1,4,3C. 1,2,4,3D. 2,1,3,4 |
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Answer» From the knowledge of theory, we know 2 represents angle of incidence 1 represents angle of emergence 4 represents angle of refraction, and (3) represents lateral diplacement. |
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| 80. |
If rafractive index of glass w.r.t. air is `3//2`, what is the refractive index of air w.r.t. glass? |
| Answer» As `.^(a)n_(g) = 3//2`, `.^(g)n_(a) =(1)/(.^(a)n_(g)) = (1)/(3//2) =(2)/(3)` | |
| 81. |
An object is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position and nature of the image. |
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Answer» Here, `u= -20cm, R =30 cm, v=?` From `(1)/(v) + 1/u =(1)/(f) =2/R, (1)/(v) =2/R - 1/u = 2/30 + 1/20= 4+3/60=7/60` `v=60/7=8.57 cm`. |
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| 82. |
Radius of curvature of a concave mirror is 25 cm. What is its focal length? |
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Answer» Correct Answer - 12.5 cm Here, `R = 25 cm, f = ?` From `f = R/2 =25/2 = 12.5 cm` |
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| 83. |
What is the relation between focal length and radius of curvature of a concave mirror? Does the same relation hold for a covcave mirror? |
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Answer» `f=1/2 R` Yes, the same relation holds for a convex mirror too. |
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| 84. |
In problem 2, what is the refractive index of medium 1 w.r.t. medium 2 ? |
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Answer» Here, `.^(2)n_(1) = ?` `.^(2)n_(1) = 1/(.^(2)n_(1))`, `.^(2)n_(1) = 1/(sqrt2) = (sqrt2)/(sqrt(2).sqrt(2)) = (1.414)/2 = 0.707` |
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| 85. |
Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in figure. A. B. C. D. |
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Answer» Correct Answer - D The ray of light incident on a concave mirror in a direction parallel to principal axis must pass through focus F, on reflection from the mirror. Fig D is the correct choice. |
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| 86. |
Define absolute refractive index of a medium. Find its value for glass in which speed of light is `2 xx 10^(8)m//s`. |
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Answer» Absolute refraactive index of a medium, `n = ("velocity of light vacuum" (c ))/("velocity of light in the medium" (v))` For glass, `n=(3 xx 10^(8) m//s)/(2 xx 10^(8) m//s) =1.5`. |
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| 87. |
The focal length of a convex mirror is 30cm. What is the distance of its centre of curvature from its focus? |
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Answer» Here,`f=30 cm, R= 2f = 2xx 3 =60 cm`. `PF= 30cm , PC=60 cm` `FC =PC -PF =60-30=30cm`. |
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| 88. |
A ray of light passing through focus of a concave mirror becomes parallel to the principal axis of the mirror on reflection. Comment. Is the reverse true ? |
| Answer» Yes, it is true. The reverse is also true. | |
| 89. |
What is the nature of image formed when an object is held at a distance of 10 cm from the pole of a concave mirror of focal length 15 cm? |
| Answer» The object lies between pole and principal focus of concave mirror. Therefore, the image formed is virtual, erect and magnified. | |
| 90. |
Draw ray diagrams showing the image formation by a concave mirror when an object is placed (a) between pole and focus of the mirror , (b) between focus and centre of curvature of the mirror (c ) at centre of curvature of the mirror , (d) a little beyond centre of curvature of the mirror (e) at infinity |
| Answer» The image will be formed at the back of concave mirror. It will be virtual, erect and magnified. | |
| 91. |
Can you change focal length of a given spherical mirror by changing the object distance from the mirror? |
| Answer» No, we cannot change focal length of a spherical mirror by changing the object distance (u), By changing u:v will change, but f will remain the same. | |
| 92. |
(a) State the relation between object distance, image distance and focal length of a spherical mirror. (b) Draw a ray diagram to show the image formed by a concave mirror when an object is placed between pole and focus of the mirror. (c ) A concave mirror of focal length 15 cm forms an image of an object kept at a distance of 10 cm from the mirror. Find the position, nature and size of the image formed by it. |
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Answer» (a) `(1)/(v)+1/u = (1)/(f)` (b) N/A (c ) Here, `f =- 15cm, u= -10cm, v=?` From `(1)/(v) + (1)/(u) = (1)/(f), (1)/(v) = (1)/(f) - (1)/(u) = (1)/(-15) + (1)/(10) = (-2+3)/(30) = (1)/(30)` or `v=30cm` i.e, image is formed at `30 cm` from the mirror, As v is + , image is virtual and erect. From `m= (-v)/(u) m = (-30)/(-10) =3`. Therefore, size of image is three times the size of object. |
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| 93. |
At what distance should an object be placed from a lens of focal length 25 cm to obtain an image on a screen placed at a distance of `50 cm` from the lens ? What will be the magnification produced in this case? |
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Answer» Here, `u=?, f=25 cm, v=50cm, m=?` From `(1)/(v) - 1/u = (1)/(f), - 1/u = (1)/(f) - (1)/(v) = 1/25 - 1/50 = 1/50` `u = -50cm`. And `m = v/u = 50/-50 = -1` Negative sign of m indicates that image would be inverted. |
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| 94. |
A doctor has prescribed lens of power `+ 1.5 D`. Find the focal length of the lens. Is the prescribed lens diverging or converging ? |
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Answer» Here, `P = + 1.5 D, f = ?` From `f = 100/P, f = 100/(1.5) = 66.7 cm` The prescribed lens is converging or convex as its power is positive. |
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| 95. |
Where should an object be placed in order of to use a convex lens as a magnifying glass? |
| Answer» A convex lens acts as a magnifying glass, when it forms a virtual, erect and magnified image of an object. For this, the object is to be placed between principal focus and optical cantre of the lens. | |
| 96. |
For what position of an object a real, diminshed image is formed by a convex lens? |
| Answer» A convex lens forms a real, inverted and diminished image when the object is placed at a distance more than twice the focal length of the lens. | |