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1)) no of image =360/given angle - 1
a)) no of image =360/90-1 = 4-1=3
b)) no of images =360/60-1 = 6-1=5
2)) i )) in case of asymmetrical
no of image = 360/50 = appox 7 so Ans:-7
ii )) in case of symmetrical
no of image = 360/50-1 = 7.2-1 = 6.2 approx 6 , so Ans:-6
a)) no of image =360/90-1 = 4-1=3
b)) no of images =360/60-1 = 6-1=5
2)) i )) in case of asymmetrical
no of image = 360/50 = appox 7 so Ans:-7
ii )) in case of symmetrical
no of image = 360/50-1 = 7.2-1 = 6.2 approx 6 , so Ans:-6
Answered by
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⭐⭐⭐⭐⭐ ANSWER ⭐⭐⭐⭐⭐
==================================
*⃣ BOOK :- CONCISE ICSE PHYSICS
*⃣ EXERCISE :- 7 (B)
*⃣ CHAPTER :- REFLECTION OF LIGHT
*⃣NUMERICALS
==================================
1] (i) Here, theta = 90°
Therefore, number of images (n)
= (360/90)-1
= 4-1
= 3 (Ans.)
==================================
(ii) Here theta = 60°
Therefore, number of images (n)
= (360/60)-1
= 6-1
= 5 (Ans.)
==================================
2] Given, theta = 50°.
n = 360/50 = 7.2 or 7 (accuracy).
(i) No. Of images formed when the object is placed asymmetrically,
n = 7 (Ans.)
(ii) No. Of images formed when the object is placed symmetrically = n-1 = 7-1 = 6 (Ans.)
==================================
⭐⭐⭐ ALWAYS BE BRAINLY ⭐⭐⭐
==================================
⭐⭐⭐⭐⭐ ANSWER ⭐⭐⭐⭐⭐
==================================
*⃣ BOOK :- CONCISE ICSE PHYSICS
*⃣ EXERCISE :- 7 (B)
*⃣ CHAPTER :- REFLECTION OF LIGHT
*⃣NUMERICALS
==================================
1] (i) Here, theta = 90°
Therefore, number of images (n)
= (360/90)-1
= 4-1
= 3 (Ans.)
==================================
(ii) Here theta = 60°
Therefore, number of images (n)
= (360/60)-1
= 6-1
= 5 (Ans.)
==================================
2] Given, theta = 50°.
n = 360/50 = 7.2 or 7 (accuracy).
(i) No. Of images formed when the object is placed asymmetrically,
n = 7 (Ans.)
(ii) No. Of images formed when the object is placed symmetrically = n-1 = 7-1 = 6 (Ans.)
==================================
⭐⭐⭐ ALWAYS BE BRAINLY ⭐⭐⭐
==================================
Anonymous:
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