Ray Optics

Let's determine the maximum angle of incidence, , for which the ray can achieve total internal reflection at the boundary between two media with refractive indices and , where .Using Snell's law, we have: For Air ( ) to : Rearrange to find : For to :Assuming is the critical angle of incidence on , and is complementary to , we use: This gives: Thus, , and using the Pythagorean identity: Substitute into Equation i: This simplifies to: This final expression gives us the maximum angle of incidence for total internal reflection at the boundary.

When a convex lens with a refractive index is placed in a medium with a higher refractive index (where ), it no longer acts as a converging lens. Instead, it behaves as a diverging lens. This happens because light rays bend away from the normal when entering the lens and towards the normal when exiting, causing the rays to spread out rather than converge. Consequently, the lens's focal length becomes negative, indicating a diverging effect.

The scenario involves an object positioned in front of a convex lens, yielding a real, inverted image that is double the size of the object. This setup implies a magnification .Initial Setup:The magnification is given by the relation: This implies: Change in Position:When the object is moved closer to the lens by a distance , the image shifts by a distance further from the lens. The new object and image distances are: Lens Formula:The lens formula is: We equate the initial and final conditions: Substituting these values, we get: Simplifying:On cross-multiplying: Solving for : Final Magnification Calculation:To find the new magnification : Calculating the absolute value: Thus, the magnitude of the magnification in the final setup is 4.

As we can see from the diagram,$ $$ a b \theta_1 \theta_2 \theta_1 i_2= \theta_1=30^{\circ} \theta_2 $$ $