Answer any one question. a. Define lateral shift and derive an expression for it due to a parallel edged glass slab. b. Explain the construction and working of a compound microscope with final image…

a. Lateral Shift and Derivation

Definition: Lateral shift is the perpendicular displacement of the emergent ray from the path of the incident ray when light passes through a parallel-edged glass slab.

Derivation: Consider a ray of light incident at angle $i$ on a glass slab of thickness $t$ and refractive index $\mu$. Let $r$ be the angle of refraction inside the slab.

1. Apply Snell’s Law at the first surface: $$\mu =\frac{\sin i}{\sin r}$$

2. Lateral displacement ($d$) is the horizontal shift between the incident and emergent rays. From the geometry of the slab: $$d=t\cdot \sin (i-r)$$

3. Express $\sin (i-r)$ using trigonometric identity: $$\sin (i-r)=\sin i\cos r-\cos i\sin r$$

4. Substitute $\sin r=\frac{\sin i}{\mu }$ and $\cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{1-\frac{{\sin }^{2}i}{{\mu }^2}}$: $$d=t\left(\sin i\sqrt{1-\frac{{\sin }^{2}i}{{\mu }^2}}-\cos i\cdot \frac{\sin i}{\mu }\right)$$ Simplifying: $$d=t\sin i\left(\sqrt{1-\frac{{\sin }^{2}i}{{\mu }^2}}-\frac{\cos i}{\mu }\right)$$

Final Expression: $$d=\frac{t\sin i}{\mu }\left(\sqrt{{\mu }^2-{\sin }^{2}i}-\cos i\right)$$

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b. Compound Microscope and Magnifying Power

Construction: A compound microscope consists of two convex lenses:

1. Objective lens (short focal length $f_1$) forms a real, inverted, and magnified image.
2. Eyepiece lens (long focal length $f_2$) further magnifies this image to form a virtual image at the near point ($D=25$ cm).

Working:

• The object is placed just beyond $f_1$ of the objective lens, producing a real image inside the tube.
• This real image acts as the object for the eyepiece, which forms a final virtual image at the near point.

Derivation of Magnifying Power ($M$):

1. Magnification by objective ($m_1$): $$m_1=\frac{v_1}{u_1}\approx \frac{v_1}{f_1}(\text{since }{u}_1\approx -f_1)$$ For small angles, $v_1\approx f_1$, so $m_1\approx \frac{v_1}{f_1}$.

2. Magnification by eyepiece ($m_2$): The eyepiece forms the final image at the near point ($D$): $$m_2=1+\frac{D}{f_2}$$

3. Total magnifying power ($M$): $$M=m_1\cdot m_2=\frac{v_1}{f_1}\left(1+\frac{D}{f_2}\right)$$ If the image formed by the objective is at the least distance of distinct vision ($D$): $$M=\frac{D}{f_1}\left(1+\frac{D}{f_2}\right)$$ For a relaxed eye (final image at infinity): $$M=\frac{v_1}{f_1}\cdot \frac{D}{f_2}$$