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A point object is placed at the centre of a glass sphere of radius 6cm and refractive index 1.5. The distance of virtual image from the surface is

A

6cm

B

4cm

C

12cm

D

9cm

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To solve the problem of finding the distance of the virtual image from the surface of a glass sphere with a point object placed at its center, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Information**: - Radius of the glass sphere (R) = 6 cm - Refractive index of glass (n) = 1.5 - The object is placed at the center of the sphere. 2. **Understand the Refraction Process**: - The point object is located at the center of the sphere. When light rays emanate from the object, they will strike the inner surface of the sphere. 3. **Use the Concept of Refraction**: - Since the object is at the center, the light rays will strike the surface at normal incidence (perpendicular to the surface). At normal incidence, the light rays do not bend; they continue in a straight line. 4. **Determine the Image Formation**: - Because the object is at the center and the light rays are not refracted, the image will form at the same location as the object, which is at the center of the sphere. 5. **Calculate the Distance from the Surface**: - The distance of the virtual image from the surface of the sphere can be calculated as follows: - The center of the sphere is 6 cm away from the surface (the radius of the sphere). - Since the image is formed at the center, the distance of the virtual image from the surface is equal to the radius of the sphere. - Therefore, the distance of the virtual image from the surface = 6 cm. ### Final Answer: The distance of the virtual image from the surface of the glass sphere is **6 cm**. ---

To solve the problem of finding the distance of the virtual image from the surface of a glass sphere with a point object placed at its center, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Information**: - Radius of the glass sphere (R) = 6 cm - Refractive index of glass (n) = 1.5 - The object is placed at the center of the sphere. ...
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