# Reflection of Light Exe-7A Numericals Concise Physics ICSE Class-9 Selina Publishers

Reflection of Light Exe-7A Numericals Concise Physics ICSE Class-9 Selina Publishers. There is the solutions of Numerical Questions of your latest textbook which is applicable in 2024-25 academic sessionVisit official Website CISCE for detail information about ICSE Board Class-9.

## Reflection of Light Exe-7A Numericals

Concise Physics ICSE Class-9 Selina Publishers

 Board ICSE Class 9 Subject Physics Writer / Publication Concise Selina Publishers Chapter-7 Reflection of Light Exe-7A Laws of Reflection Topics Solution of Numerical Academic Session 2024-2025

### Exercise- 7A Laws of Reflection

page -170

#### Numericals :

##### Que: 1- A ray is incident on a plane mirror . Its reflected ray is perpendicular to the incident ray . Find the angle of incidence .

Solution- Angle of incidence (i) + Angle of reflection(r) = 90But, as per the laws of reflection, i = r

Therefore, 2 i = 90o

Or, i = r = 45o

##### Que: 2- A man standing in front of a plane mirror finds his image at a distance 6 metre from himself. What is the distance of man from the mirror?

Solution- Distance between man and his image = 6mDistance between man and mirror + distance between mirror and image = 6m

But, Distance between man and mirror (object distance) = distance between mirror and image (image distance)

Therefore, distance of man from mirror = 6/2 =3m.

##### Que:3 –An insect is sitting in front of a plane mirror at a distance 1 m from it.

(a) Where is the image of the insect formed?

(b) What is the distance between the insect and its image?

Solution- (a) Image of the insect is formed 1m behind the mirror.
(b) Distance between the insect and the man = 2m.

##### Que: 4- An object is kept at 60 cm in front of a plane mirror. If the mirror is now moved 25 cm away from the object, how does the image shift from its previous position?

Solution- Initially, distance of the object from the mirror = 60 cm. Therefore, image is formed at a distance 60 cm from the mirror, behind it.

Thus, initial distance between the object and image = 60 + 60 = 120 cm

If the mirror is moved 25 cm away from the object,

The new distance of the object from the mirror = 60 + 25 = 85 cm

The new image is now at a distance 85 cm from the mirror behind it.

Thus, new distance of the image from the object = 85 + 85 = 170 cm

Taking the position of the object as reference point, the distance between the two positions of the image = new distance of image from the object – initial distance of the image from the object

= (170 – 120) cm = 50 cm

Thus, the image shifts 50cm away.

##### Que: 5- An optician while testing the eyes of a patient keeps a chart of letters 3 m behind the patient and asks him to see the letters on the image of chart formed in a plane mirror kept at distance 2 m in front of him. At what distance is the chart seen by the patient?

Solution- Distance between man and chart = 3m

Distance between man and mirror = 2m

Therefore, distance between chart and mirror = 5 m

Now, final image is formed on the mirror, which is at a distance of 2 m from the man, therefore, the chart as seen by patient is (5m + 2m =) 7m away.

— : end of Reflection of Light Exe-7A Numericals Concise Physics ICSE Class-9 Selina Publishers :- –

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