Compound Microscope:
A compound microscope consists of two lenses: the objective lens and the eyepiece. The objective lens forms a real, inverted, and magnified image of the object, and the eyepiece magnifies this image further.
- The object is placed slightly inside the focal point of the objective lens.
- The objective lens produces a real, inverted, and magnified image, which acts as the object for the eyepiece.
- The eyepiece forms an image at infinity, as the final image is formed at infinity (in a relaxed eye condition).
The total magnification \( M_{\text{total}} \) of the microscope is the product of the magnification produced by the objective lens \( M_{\text{obj}} \) and the magnification produced by the eyepiece \( M_{\text{eyepiece}} \):
\[ M_{\text{total}} = M_{\text{obj}} \times M_{\text{eyepiece}} \]
The magnification of the objective lens is given by:
\[ M_{\text{obj}} = - \frac{v_{\text{obj}}}{u_{\text{obj}}} \]
Where \( v_{\text{obj}} \) is the image distance and \( u_{\text{obj}} \) is the object distance for the objective lens.
The magnification of the eyepiece is given by:
\[ M_{\text{eyepiece}} = \frac{D}{f_{\text{eyepiece}}} \]
Where \( D \) is the near point distance of the eye (usually taken as 25 cm), and \( f_{\text{eyepiece}} \) is the focal length of the eyepiece.
Thus, the total magnification is the product of these two magnifications.
Simar, Tanvi, and Umara were partners in a firm sharing profits and losses in the ratio of 5 : 6 : 9. On 31st March, 2024, their Balance Sheet was as follows:
Liabilities | Amount (₹) | Assets | Amount (₹) |
Capitals: | Fixed Assets | 25,00,000 | |
Simar | 13,00,000 | Stock | 10,00,000 |
Tanvi | 12,00,000 | Debtors | 8,00,000 |
Umara | 14,00,000 | Cash | 7,00,000 |
General Reserve | 7,00,000 | Profit and Loss A/c | 2,00,000 |
Trade Payables | 6,00,000 | ||
Total | 52,00,000 | Total | 52,00,000 |
Umara died on 30th June, 2024. The partnership deed provided for the following on the death of a partner:
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is: