Question

By drawing ray-diagram, explain the formation of image by a compound microscope. Establish the formula for its magnifying power.
OR
What is meant by binding energy of nucleus? Draw the graph of binding energy per nucleon with respect to the mass number. Discuss fission and fusion reactions with the help of this curve.

Hint:

Microscope: magnifying power is the product of objective magnification \(L/f_o\) and eyepiece magnification \((1+D/f_e)\). Binding energy: \(B.E.=\Delta m\,c^2\); both fission and fusion move nuclei toward the peak of the \(B.E./A\) curve, releasing energy.

Answer 1

OPTION 1: Compound Microscope

Step 1: Construction. A compound microscope uses two convex lenses mounted at the ends of a tube. The lens facing the object is the objective (small focal length \( f_o \) and small aperture); the lens near the eye is the eyepiece (focal length \( f_e \), with \( f_e > f_o \)).

Step 2: Formation of image (ray diagram in words). The tiny object \( AB \) is placed just outside the principal focus of the objective (between \( F_o \) and \( 2F_o \)). Rays from \( AB \) refract through the objective and form a real, inverted and magnified image \( A'B' \) on the other side of the objective, inside the tube. This intermediate image \( A'B' \) falls just inside the focal length of the eyepiece. The eyepiece now works like a simple magnifier and forms a final image \( A''B'' \) which is virtual, still more magnified, and inverted with respect to the object. The eyepiece position is adjusted so that \( A''B'' \) is formed at the least distance of distinct vision \( D \) (= 25 cm).

Step 3: Definition of magnifying power. Magnifying power \( M \) = (angle \( \beta \) subtended at the eye by the final image) \( / \) (angle \( \alpha \) subtended at the eye by the object when it is kept at distance \( D \)). It equals the product of the linear magnification of the objective and the angular magnification of the eyepiece:
\[ M = m_o \times m_e \]

Step 4: Magnification of the objective. If \( u_o \) and \( v_o \) are the object and image distances for the objective,
\[ m_o = \frac{h'}{h} = \frac{v_o}{u_o} \approx \frac{L}{f_o} \]
where \( L \) is the tube length (distance between the two lenses) and the object lies almost at the focus, so \( u_o \approx f_o \) and \( v_o \approx L \).

Step 5: Magnification of the eyepiece. The eyepiece acts as a simple microscope. For the final image at the near point \( D \),
\[ m_e = 1 + \frac{D}{f_e} \]

Step 6: Total magnifying power. Multiplying,
\[ M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right) \]
When the final image is formed at infinity (relaxed eye), \( m_e = D/f_e \), so
\[ M = \frac{L}{f_o}\cdot\frac{D}{f_e} \]
Because \( f_o \) and \( f_e \) are both small, \( M \) is very large.
\[\boxed{M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)}\]

OPTION 2: Binding Energy of Nucleus

Step 1: Meaning of binding energy. The measured mass of a nucleus is always less than the total mass of its free protons and neutrons. This missing mass is the mass defect:
\[ \Delta m = \big[\,Z\,m_p + (A-Z)\,m_n\,\big] - M_{nucleus} \]
where \( Z \) = number of protons, \( A \) = mass number, \( m_p, m_n \) are proton and neutron masses. The binding energy is the energy equivalent of this mass defect. It is the energy needed to split the nucleus completely into free nucleons, or the energy released when free nucleons combine to form the nucleus.
\[ B.E. = \Delta m\, c^2 \]

Step 2: Binding energy per nucleon. Dividing the binding energy by the mass number gives
\[ \frac{B.E.}{A} \]
This measures the average stability of each nucleon. A larger value means a more tightly bound, more stable nucleus.

Step 3: The curve. When \( B.E./A \) (in MeV) is plotted against mass number \( A \): (a) it rises steeply for light nuclei; (b) it reaches a broad maximum of about \( 8.8 \) MeV near \( A \approx 56 \) (iron), and stays nearly flat (about \( 8.5 \) MeV) for \( 30 < A < 170 \); (c) it then falls slowly to about \( 7.6 \) MeV for very heavy nuclei such as uranium (\( A \approx 238 \)). Nuclei near the peak are the most stable.

Step 4: Nuclear fission. A very heavy nucleus (like uranium) has a smaller \( B.E./A \) (about \( 7.6 \) MeV). When it splits into two medium-sized nuclei that lie nearer the peak (\( B.E./A \approx 8.5 \) MeV), the products are more tightly bound. The extra binding energy is released, about \( 200 \) MeV per fission.

Step 5: Nuclear fusion. Very light nuclei (like hydrogen isotopes) have very small \( B.E./A \). When they combine (fuse) into a heavier nucleus that has a much larger \( B.E./A \), energy is again released. This powers the Sun and stars.

In both processes the reaction moves nucleons toward higher \( B.E./A \), i.e. toward the peak of the curve, and the increase in binding energy appears as released energy.
\[\boxed{B.E. = \Delta m\,c^2,\quad \text{energy released} \propto \text{rise in } B.E./A}\]