An object is located 3s m in front of a plane mirror, the image formed by the mi

Question

An object is located 3s m in front of a plane mirror, the image formed by the mirror appears to be 3.8 m in front of the mirror 1.9 m in front of the mirror on the mirror's surface 3.8 m behind front of the mirror 1.9 m behind front of the mirror the optical fiber structure and mechanism of light transmission is shown in the figure. What condition does the incident angle (theta_0) have to meet for the light being able to propagate in the optic fiber? The speed of light in a certain medium is 2.2 times 10^8 m/s. What is the index of refraction of this medium? 1.0 1.6 104 0.7 0.8 If you stand in front of a convex mirror at the same distance from it as its focal length you want see your Image because its focused at different distance. You won't see your image because there is none A beam off light passes through two ideal polarizers. In this process, the intensity of the light transmitted is What is the angle between the transmission axes of the two polarizers? 60 degree 22 degree 22 degree 18 degree in calculating the magnification from a lens, a negative result means the image is directed downward the image is inverted. The calculation is incorrect the image is large than the object. The image is small than the object.

Explanation / Answer

here,

A plane mirror makes an image of objects in front of it; these images appear to be behind the plane in which the mirror lies

the object distance is 3.8 m in front of the mirror

so, the image formed is

d) 3.8 m behind the mirror's surface

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