WBCHSE Class 12 Physics Reflection Of Light Short Question And Answers

Reflection Of Light Short Question And Answers

Question 1. Use the mirror equation to deduce that:

1. An object placed between ƒ and 2f of a concave mirror produces a real image beyond 2f
2. The virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole
3. A convex mirror always produces a virtual image independent of the location of the object
4. An object placed between the pole and the focus of a concave mirror produces a virtual and enlarged image.

Answer:

1.  For a concave mirror, the mirror formula is given by:

⇒ \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

The object lies between ƒ and 2f,

2f>u>f Or, \(\frac{1}{2 f}<\frac{1}{u}<\frac{1}{f}\)

Or, \(\frac{1}{2 f}-\frac{1}{f}<\frac{1}{u}-\frac{1}{f}<\frac{1}{f}-\frac{1}{f}\)

Or, \(-\frac{1}{2 f}<-\left(\frac{1}{f}-\frac{1}{w}\right)\)<0

∴ 0>v>2f

Since v is -ve, the image is real and lies beyond 2f.

2. In the case of a convex mirror:

⇒ \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)………………..(1)

Or, \(\frac{u}{v}=\frac{u}{f}-1\)

Or, \(-\frac{1}{m}=\frac{u-f}{f}\)

Or, – m\(=\frac{f}{u-f}\)

∴ – m = \(\frac{f}{f-u}\)

∴ In this case u is -ve and f is +ve ,f-u>f.

∴ m<1 i.e the image is diminished

From equation (1), v= \(\frac{u f}{u-f}\)

∴ When u= 0 , v= 0 and when  u = ∞, v= f [substituting equation (1)]

∴ The image lies between the pole and the focus

3. For a convex mirror, \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\):

2. In this case ƒ is always +ve and u is always negative

∴ v is always +ve i.e., the image is virtual.

4. In case of a concave mirror, u and ƒ are both negative.

The general equation of mirror is

⇒ \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)

When the object lies between the pole and the focus,

f<u<0 or,> 0 \(\frac{1}{f}-\frac{1}{u}\)>0

∴ \(\frac{1}{v}>0\)  [using the mirror equation]

i.e., v is positive and lies to the right pole,

Since v > |u|, m is positive and greater than 1. So, the image is enlarged.

WBBSE Class 12 Reflection of Light Q&A

Question 2. You have learned that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
Answer:

When the incident beam is convergent, then the image formed by a plane or convex mirror produces a real image.

In both cases the convergent beam would meet at P in the absence of the mirrors i.e., P is the virtual object, and P’ is the real image of P.

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Question 3.  Light incident normally on a plane mirror attached to a galvanometer coil retraces backward as 1.55. A current in the coil produces a deflection of 3.5° of the
Answer:

For this reason, paraboloidal mirrors are used in car headlights and searchlights instead of spherical concave mirrors.

What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

The reflected ray rotates through an angle 20 for a rotation of the mirror and the deviation of the spot of light on the scale is d (say).

∴ \(\frac{d}{A B}\) = tan2θ

∴ d = 1.5× tan 7°

= 1.5 × 0.122

= 0.183m

= 18.3 cm

v = f = 110 mm

For the secondary mirror,

Short Questions on Laws of Reflection

Question 4. A Cassegrain telescope is built with mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, then where will the final image of an object at infinity be?
Answer:

The image formed by the primary mirror is the virtual object for the secondary mirror.

For the primary mirror , u = ∞, f = \(\frac{220}{2}\)

= 110 mm

∴ v= f = 110 mm

For the secondary mirror

f= \(\frac{140}{2}\)

= 70 mm

And u= 110- 20 = 90 mm

using mirror formula \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) we get

⇒ \(\frac{1}{v}=\frac{1}{70}-\frac{1}{90}\)

Or, v= 315mm

∴ The image will be formed at a distance of 315 mm on the right of the secondary mirror.

Question 5. At what position should an object be placed in front of a spherical mirror such that the size of the image is equal to that of the object?
Answer:

The spherical mirror has to be a concave mirror and the object should be placed at the center of curvature for the size of the image to be equal to that of the object.

Question 6.

1. Under what condition will the object and image always be on the same side of the focus of a concave mirror?
Answer:

Objects placed at focus or beyond the focus of a concave mirror, the image will always be formed at the same side of the focus.

Common Short Questions on Spherical Mirrors

2. An image of size = \(\frac{1}{n}\) times the object size is formed in a convex mirror. If r is the radius of curvature of the mirror, what would be the object distance?
Answer: 

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{2}{r}\)

Or, \(\frac{u}{\nu}+1=\frac{2 u}{r}\)

Or, \(\frac{u}{v}=\frac{2 u}{r}-1\)

= \(\frac{2 u-r}{r}\)

∴ Magnification, m= – \(\frac{v}{u}=\frac{r}{r-2 u}\)

According to the question , m= \(\frac{1}{n}\)

n= \(\frac{r-2 u}{r}\)

Or, nr= r- 2u

Or, 2u – r- nr

Or , u= \(\frac{(1-n) r}{2}\)

Thus, u = \(-\frac{(n-1) r}{2}\).

Practice Short Questions on Image Formation by Mirrors

Question 7. An object AB is kept in front of a concave mirror as shown in the 

1. Compare the ray diagram showing the image formation of the object.

Answer:

The ray diagram and the Image formation are

2. How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black?
Answer:  The position will remain unchanged, but image intensity will reduce.