The figure below shows a thin converging lens for which the radii of curvature of its surfaces have magnitudes of 8.99 cm and 11.1 cm. The lens is in front of a concave spherical mirror with the radius of curvature R = 8.38 cm. Assume the focal points F, and F₂ of the lens are 5.00 cm from the center of the lens. F₁ F₂ (a) Determine the index of refraction of the lens material. (b) The lens and mirror are 20.0 cm apart, and an object placed. 14 cm to the left of the lens. Determine the position (in cm) of the final image. (Give its magnitude and location with respect to the lens.) distance cm location to the left of the lens ✔✔✔ (c) Determine its magnification as seen by the eye in the figure. (d) Is the final image inverted or upright? inverted O upright Q (e) For what initial object distance (in cm) would the final image form at exactly the same location as the object? cm What would be the magnification of the image in this case? M=

We know that the lens maker formula is given as: So, for the given focal length, the refractive index of the lens can be determined.Substituting the given values, we get Here, is taken positive as it lies in the outgoing path of the ray of light which is according to sign convention. And, since lies on the incident side of the ray of light.To determine the position of the resulting image, we have to find the image due to the lens and use this image as the real object for the mirror. The result of these two processes gives us the final image.Using the lens equation, we have:where, i=image distance, f= focal length, o=object distanceThis implies that the image formed is to the right side of the lens and is a real image.Magnification by the lens is,Now, from the geometry, we observe that:Using the mirror formula to find the distance of the image formed by the mirror. We have,,where is the object distance from the mirror, is the image distance from the mirror and R is the radius of curvature of the mirror., which symbolizes that the image formed is virtual.Magnification by the mirror is,Final magnification is. The final image is inverted as the value of the magnification is negative.(a) The refractive index of lens material is 1.99(b) Final ImageDistance= 21.82 cmLocation = to the left of the lens.(c) The magnification of the final image is -4.90(d) The final image is inverted.(According to guidelines, the first 3 questions have to be solved.)