## Saturday, December 10, 2011

### Class 10 - Physics - Light - Numerical and Questions

(See also at set 1)

Q1: When an object is placed at a distance of 60 cm from a convex mirror, the magnification produced is 1/2.
Where should the object be placed to get a magnification of 1/3?

Answer:

Q2: When an object is kept at a distance of 60cm from a concave mirror, the magnification is 1/2. Where should the object be placed to get magnification of 1/3?

Answer: (Hint: steps are same as in the above question, except it is concave mirror, m = -1/2. Compute f in first case i.e. f = -20 cm and the compute u in case II i.e. u = 80 cm)

Q3(CBSE Board): A concave lens made of material of refractive index (n1) is kept in a medium of refractive index (n2). A parallel beam of light is incident on the lens. Complete the path of the rays of light emerging from a concave lens if:
(a) n1 > n2
(b) n1 = n2
(c) n1 < n2

Answer:

Q4: A convex lens made of material of refractive index (n1) is kept in a medium of refractive index (n2). A parallel beam of light is incident on the lens. Complete the path of the rays of light emerging from a concave lens if:
(a) n1 > n2
(b) n1 = n2
(c) n1 < n2
Answer:
Q5: Which of the two has a great power? A lens of short focal length or a lens of large focal length?
Answer: lens of short focal length.
Q6: When a convex lens and a concave lens of equal focal lengths are placed side by side, what will be the equivalent power and the combined focal length.
Answer: Equivalent Power = 0 and Combined focal length = infinity.

Q7: When a plane mirror is rotated by an angle $\theta$, what will be the angle of rotation of the reflected ray for the fixed incident ray?

Answer: The reflected ray will be rotated twice w.r.t to the angle of rotation of mirror. i.e. $2\theta$

Consider the following figure:

As shown, mirror M is rotated by an angle $\phi$. Let I be incident ray, N be normal and R is reflected ray.
Before rotation, let the incident angle be $\theta$
After rotation, normal moves by angle $\phi$,
$\therefore \text{incident angle } \angle {ION'} = \theta + \phi \newline \angle{ROR'} = \angle{IOR'} - \angle{IOR} = 2(\theta + \phi) - 2\theta = 2\phi$

Q8: With respect to air the refractive index of kerosene is 1.44 and that of diamond is 2.42. Calculate the refractive index of diamond with respect to kerosene.

Answer:

#### 16 comments:

1. it is much helpful

1. it's very helpful not much helpful use correct english

2. Hmmm. Even u forgot to put a comma in your sentence. Use proper punctuation.ðŸ˜‘ðŸ˜‘ðŸ˜¡

3. Need to give more. Numericals

4. @TanishaSharma, a sentence always starts with a capital letter.
@HARSHVARDHANSINGHChauhan, it’s not “u” it is “you” and it’s not “even u” it’s “you even”.

2. In Q3 the last n1<n2 should'nt be a convex lens???

1. they have said "if" so we can accept that

3. i cud hav lookd it earlier

1. This comment has been removed by the author.

4. In Q2, u cannot be (+)80.... u has to be -ive

1. this is very true but it is said that in hint apply that u is if positive so what will be the answer u should read the question nicely

5. Questions are too easy........post sumthing intresting

6. Thank you very much. Found this post very useful

7. thanks it is very useful for me!