By drawing a ray diagram, explain the formation of image in a compound microscope. Establish the formula for magnifying power for it.

Ray diagram and Explanation: In a compound microscope, an objective lens and an eyepiece are used. The object is placed at a distance between the focal point and the objective lens. The image formed by the objective is real and inverted, and it acts as the object for the eyepiece. The eyepiece forms a virtual, magnified image of the real image created by the objective. - Magnifying power formula: The magnifying power \( M \) of the compound microscope is the product of the magnification by the objective lens \( M_1 \) and the magnification by the eyepiece \( M_2 \): \[ M = M_1 \times M_2 = \frac{D}{f_o} \times \frac{L}{f_e}, \] where: - \( f_o \) is the focal length of the objective lens, - \( f_e \) is the focal length of the eyepiece, - \( L \) is the length of the microscope tube, - \( D \) is the near point of the eye (usually taken as 25 cm for normal vision).
The path of scattered \( \alpha \)-particle is:
The maximum focal length of convex lens is for:
The power consumed in alternating current in a circuit containing only a capacitor will be:
The path of scattered \( \alpha \)-particle is:
What is Bohr's quantum condition postulate? How is it explained by de Broglie? What are the shortcomings of Bohr's atomic model?
An electromagnetic wave propagating through vacuum, described by \( E = E_0 \sin(kx - \omega t) \), \( B = B_0 \sin(kx - \omega t) \), then: