25 + Interview Questions in RAY-OPTICAL-AND-INSTRUMENTS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification.
Answer» `O = 4.5 cm, u = -12cm, f = 15 cm` `(1)/(u)+(1)/(upsilon)=(1)/(f) rArr (-1)/(12)+(1)/(upsilon)=(1)/(15)` `(1)/(v)=(1)/(12)+(1)/(15)=(5+4)/(60)=(9)/(60)=(3)/(20)` `upsilon=(20)/(3)=6.7cm` Image is virtual and erect and is formed behind the mirror. `(I)/(O)=(-v)/(u) rArr (1)/(4.5)=(-20)/(3xx(-12))` `I=(20xx4xx2)/(3xx12xx10)=2.5cm` As the needle is moved further from the mirror, the image moves towards the focus and gets progressively diminished in size.

Discussion

2.	An object is placed at a distance of 20 cm from a thin double convex lens of focal length 15 cm. Find the position and magnification of the image.
Answer» Given that u = 20 cm, f = 15 cm `(1)/(f)=(1)/(v)+(1)/(u)` `(1)/(v)=(1)/(f)-(1)/(u)=(1)/(15)-(1)/(20)` `(1)/(v)=(20-15)/(15xx20)=(5)/(15xx20)` `(1)/(v)=(1)/(60)` `"v = 60 cm."` `"Magnification (m) "=(-v)/(u)` `m=(-60)/(-20)(u=-20cm)` m = 3

Discussion

3.	d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease ?
Answer» The apparent depth for oblique viewing decreases from its value for nearnormal viewing.

Discussion

4.	A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out ? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer» If r is the radius of the large circle from which light comes out, C is the critical angle for water - air interface, then `tanC=(DB)/(DO)=(r)/(d)` `r = d tan C` Area of circle, `A = pir^(2)` `A=pi(d tan C)^(2)` `A=pid^(2).(sin^(2)C)/(cos^(2)C)` `A=pid^(2)(sin^(2)C)/(1-sin^(@)C)` `"But Sin C"=(1)/(mu)=(1)/(1.33)~~0.75` `A=(pi(0.8)^(2)(0.75)^(2))/(1-(0.75)^(2))=2.6m^(2)`

Discussion

5.	c) The above person prefers to remove his spectacles while reading a book. Explain why ?
Answer» c) The myopic person may have a normal near point, i.e., about 25 cm (or even less). In order to read a book with the spectacles, such a person must keep the book at a distance greater than 25 cm so that the image of the book by the concave lens is produced not closer than 25 cm. The angular size of the book ( or its image) at the greater distance is evidently less than the angular size when the book is placed at 25 cm and no spectacles are needed. Hence, the person prefers to remove the spectacles while reading.

Discussion

6.	The far point of a myopic person is 80 cm in front of the eye . What is the natuure and power of the lens required to correct the defect ?
Answer» Solving as in the previous example, we find that the person should use a concave lens of focal length `=-80cm`. i.e., of power `=-1.25` dioptres.

Discussion

7.	a) Figure shows a cross - section of a light pipe made of a glass fibre of refractive inde 1.68. The outer covering of the pipe is made of a material of refracitive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figures. b) What is the answer if there is no outer covering of the pipe?
Answer» `mu=(1.68)/(1.44)=(1)/(sinC)` `SinC=(1.44)/(1.68)=0.8571` `C=59^(@)` Total internal reflection takes place when `i gt 59^(@)` or angle r may have value between 0 to `31^(@)` `r_("max")=31^(@)` Now `(sin i_("max"))/(sin r_("max"))=1.68` `(sin i_("max"))/(sin 31^(@))=1.68` `Sin i_("max")=0.8562, i_("max")~~60^(@)` Thus all incident rays of angles in the range `0 lt i lt 60^(@)` will suffer total internal reflections in the pipe. b) If there is no outer covering of the pipe `Sin C=(1)/(mu)` `=(1)/(1.68)=0.5962` `sin C=sin 36.5^(@)` `C=36.5^(@)`

Discussion

8.	Define power of a lens and write its unit
Answer» Power of a lens : Power of a lens is defined as its bending ability and is measured as reciprocal of focal length in metre. `therefore" Power of a lens (P)"=(1)/("f (in metres)")=(100)/("f(in cms)")` `"Unit" rarr "Dioptre (D)"`

Discussion

9.	What are the laws of reflection through curved mirrors ?
Answer» i) 'The angle of reflection equals to the angle of incidence '. ii) 'The incident ray, reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane'.

Discussion

10.	What is optical density and how is it different from mass density ?
Answer» Optical density : Optical density is defined as the ratio of the speed of light in media. Mass density : Mass per unit volume is defined as mass density. Mass density of an optically denser medium less than that of optically rarer medium.

Discussion

11.	a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain.
Answer» Rays converging to a point (behind) a plane or convex mirror are reflected to a point infront of the mirror on the screen. In other words a plane or convex mirror can produce a real image if the object is virtual.

Discussion

12.	The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye ?
Answer» a) `u=-25cm, upsilon=-75cm` `1//f=1//25-1//75`, i.e., `f=37.5cm.` The corrective lens needs to have a converging power of `+2.67` dioptres.

Discussion

13.	The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye ? In what way does the corrective lens help the above person? Does the lens magnify objects held near the eye ?
Answer» b) The corrective lens produces a virtual image (at 75 cm) of an object at 25 cm. The angular size of this image is the same as that of the object. In this sense the lens does not magnify the object but merely brings the object to the near point of the hypermetric eye, which then gets focussed on the retina. However, the angular size is greater than that of the same objectat the near point (75 cm) viewed with-out the spectacles.

Discussion

14.	Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. A current in the coil produceds a deflection of `3.5^(@)` of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away ?
Answer» Here, `theta=3.5^(@)` `"x = 1.5m, d = ?"` When the mirror turns through an angle `theta`, the reflected ray turns through double the angle. `2theta=2xx3.5^(@)=7^(@)=(7pi)/(180)" rad"` from figure, `tan 2 theta=(SA)/(OS)=(d)/(1.5)xxd` `=1.5xx(7pi)/(180)m=0.18m` `d=1.5 tan 2 theta` `~~1.5 (2theta)` `=1.5xx(7pi)/(180)m=0.18m`

Discussion

15.	A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when b) the final image is formed at the least distance of distinct vision (25 cm) ?
Answer» Here, `f_(0)=140cm, f_(e)=5.0cm` Magnifying power = ? b) When final image is at the least distance of distinct vision, Magnifying power `=(-f_(0))/(f_(e))(1+(f_(e))/(d))=(-140)/(5)(1+(5)/(25))=-33.6`

Discussion

16.	A double convex lens of focal length 15 cm is used as a magnifying glass in order to produce an erect image which is 3 times magnified. What is the distance between the object and the lens ?
Answer» f = 15 cm m = 3 Magnifying power `(m)=(-upsilon)/(u)=(f)/(f-u)` `3=(15)/(15-u)` `45-3u=15` `3u=45-15` `3u=30` `u=(30)/(3)=10cm.`

Discussion

17.	A card sheet divided into squares each of size `"1 mm"^(2)` is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. a) What is the magnification in produced by the lens? How much is the area of each square in the virtual image ?
Answer» a) Here, area of each (object) square `="1 mm"^(2),` `u=-9cm, f=10cm` As `(1)/(upsilon)-(-1)/(upsilon)=(-1)/(f)` `(1)/(upsilon)=(1)/(f)+(1)/(u)=(1)/(10)(-1)/(9)=(-1)/(90)` `v=-90cm` Magnification, `m=(upsilon)/(\|u\|)=(90)/(9)=10` `therefore` Area of each square in virtual image `=(10)^(2)xx1="100 sq. mm"`

Discussion

18.	A card sheet divided into squares each of size `"1 mm"^(2)` is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. What is the angular magnification (magnifying power) of the lens ?
Answer» b) `"Magnifying power "=(d)/(u)=25//9=2.8`

Discussion

19.	Find two positions of an object, placed in front of a concave mirror of focal length 15cm, so that the image formed is 3 times the size of the object.
Answer» f = 15 cm m =3 i) `m=(-upsilon)/(u)=(f)/(f-u)` `3=(15)/(15-u)` `45-3u=15` 3u = 30 u = 10 cm. ii) `m=(f)/(u-f)` `3=(15)/(u-15)` `3u-45=15` 3u = 60 u = 20 cm.

Discussion

20.	The focal length of a concave lens is 30 cm. Where should an object be placed so that its image is `1//10` of its size ?
Answer» `"f = 30 cm, h"_(1)=h, h_(2)=(h)/(10)` `m=(-"v")/(u)=(h_(2))/(h_(1))` `(-v)/(u)=(h)/(10h) rArr (-1)/(v)=(10)/(u)` `(1)/(f)=(1)/(v)+(1)/(u) rArr (1)/(u) =(1)/(f)-(1)/(v) rArr (1)/(u)=(1)/(30)+(10)/(u)` `(1)/(u)-(10)/(u)=(1)/(30) rArr (1)/(u) (1-10)=(1)/(30) rArr (-9)/(u)=(1)/(30) rArr u = -"270 cm."`

Discussion

21.	A concave mirror of focal length 10 cm is placed at a distance 35 cm from a wall. How far from the wall should an object be placed so that its real image is formed on the wall ?
Answer» `"f = 10 cm, "upsilon " = 35 cm"` `(1)/(f)=(1)/(upsilon)+(1)/(-u)" (using sign convention)"` `(1)/(u)=(1)/(upsilon)-(1)/(f)=(1)/(35)-(1)/(10)` `(1)/(u)=(10-35)/(35xx10)=(-1)/(14)` `U = -14cm.` Distance of the object from the wall `=35-14=21cm`.

Discussion

22.	A small pin fixed on a table top is viewed from above from a adistance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table ? Refractive index of glass = 1.5. Does the answer depend on the location of the slab ?
Answer» `mu=1.5,` real thickness = 15 cm `mu=("Real depth")/("Apparent depth")` `15=("1.5")/("Apparent Depth")` `therefore" Apparent Depth "=(15)/(1.5)=10cm` `therefore" Pin appears raised by "15-10=5cm.` The result is independent of the location of the slab.

Discussion

23.	A concave mirror produces an image of a long vertical pin, placed 40 cm from the mirror, at the position of the object. Find the focal length of the mirror.
Answer» Given `u=upsilon=" 40 cm"` `(1)/(f)=(1)/(upsilon)+(1)/(u)` `(1)/(f)=(1)/(40)+(1)/(40)` `(1)/(f)=(2)/(40)` f = 20 cm.

Discussion

24.	Use the mirror equation to deduce that : d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note : The exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Answer» The mirror equation is `(1)/(u)+(1)/(v)=(1)/(f)` But `m=(upsilon)/(u)=(f)/(u-f)=(upsilon-f)/(f)` Hence `m=(f)/(u-f)` `upsilon =f(m+1)` d) When `u gt 0 lt f` we get `m=(f)/(u-f)=(f)/((gt 0 lt f)-f)=(f)/((gt -f lt 0))` `= gt -1` (`because` m is negative, image is virtual and enlarged because is numerically `y gt 1`).

Discussion

25.	Use the mirror equation to deduce that : a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. [Note : The exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Answer» The mirror equation is `(1)/(u)+(1)/(v)=(1)/(f)` But `m=(upsilon)/(u)=(f)/(u-f)=(upsilon-f)/(f)` Hence `m=(f)/(u-f)` `upsilon =f(m+1)` When `u gt f lt 2f`, we get `m=(f)/(u-f)` `=(f)/((gt f lt 2f)-f)` `=(f)/((gt O lt f))= gt 1` Hence `upsilon = f(m+1)=f(gt1+1)` or `upsilon gt 2f.` Since for concave mirror, f is negative, `upsilon` becomes negative. It means image produced is real and beyound 2f.

Discussion

Explore topic-wise InterviewSolutions in .

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification.

An object is placed at a distance of 20 cm from a thin double convex lens of focal length 15 cm. Find the position and magnification of the image.

d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease ?

A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out ? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

c) The above person prefers to remove his spectacles while reading a book. Explain why ?

The far point of a myopic person is 80 cm in front of the eye . What is the natuure and power of the lens required to correct the defect ?

Define power of a lens and write its unit

What are the laws of reflection through curved mirrors ?

What is optical density and how is it different from mass density ?

a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain.

The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye ?

The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye ? In what way does the corrective lens help the above person? Does the lens magnify objects held near the eye ?

Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. A current in the coil produceds a deflection of `3.5^(@)` of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away ?

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when b) the final image is formed at the least distance of distinct vision (25 cm) ?

A double convex lens of focal length 15 cm is used as a magnifying glass in order to produce an erect image which is 3 times magnified. What is the distance between the object and the lens ?

A card sheet divided into squares each of size `"1 mm"^(2)` is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. What is the angular magnification (magnifying power) of the lens ?

Find two positions of an object, placed in front of a concave mirror of focal length 15cm, so that the image formed is 3 times the size of the object.

The focal length of a concave lens is 30 cm. Where should an object be placed so that its image is `1//10` of its size ?

A concave mirror of focal length 10 cm is placed at a distance 35 cm from a wall. How far from the wall should an object be placed so that its real image is formed on the wall ?

A concave mirror produces an image of a long vertical pin, placed 40 cm from the mirror, at the position of the object. Find the focal length of the mirror.

Use the mirror equation to deduce that : d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note : The exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Use the mirror equation to deduce that : a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. [Note : The exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]