A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image

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  1. Class 10
  2. Chapter 10 Class 10 - Light - Reflection and Refraction (Term 1)

Transcript

Example 10.1 A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image Radius of curvature = 3 m We know that Focal length = (π‘…π‘Žπ‘‘π‘–π‘’π‘  π‘œπ‘“ π‘π‘’π‘Ÿπ‘£π‘Žπ‘‘π‘’π‘Ÿπ‘’)/2 = 3/2 = 1.5 m Since the focus of convex mirror is behind the mirror, the focal length will be positive Focal length = f = + 1.5 m Now, Bus (Object) is placed infront of the mirror ∴ Object distance will be negative Object distance = u = βˆ’5 m Finding position of image Let the image distance = v Using mirror formula, 1/𝑓 = 1/𝑣 + 1/𝑒 1/𝑓 βˆ’ 1/𝑒 = 1/𝑣 1/𝑣 = 1/𝑓 βˆ’ 1/𝑒 1/𝑣 = 1/1.5 βˆ’ 1/((βˆ’5)) 1/𝑣 = 1/1.5 + 1/5 1/𝑣 = 10/15 + 1/5 1/𝑣 = (10 + 3)/15 1/𝑣 = 13/15 1/𝑣 = 15/13 𝑣 = + 1.15 m ∴ Image is produced at a distance of 1.15 m +ve sign indicates that image is formed behind the mirror. Finding size of image Magnification = (βˆ’π‘£)/𝑒 = (βˆ’ (πΌπ‘šπ‘Žπ‘”π‘’ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’))/((𝑂𝑏𝑗𝑒𝑐𝑑 π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’)) = (βˆ’(1.15))/(βˆ’5) = 1.15/5 = 115/500 = 23/100 = + 0.23 Since the magnification is less than 1 ∴ Image is diminished and it is smaller in size by a factor of 0.23 And since magnification is positive, ∴ Image is erect Hence, Image is virtual, erect and smaller in size by a factor of 0.23

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 11 years and a teacher from the past 11 years. He teaches Science, Accounts and English at Teachoo