For an object at infinite distance, what do you substitute for 'u' in the formula $\frac{i}{f}=\frac{1}{v}-\frac{1}{u}$ ?Solution in Telugu

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If $f\left(x\right)=\log \left\{\frac{u\left(x\right)}{v\left(x\right)}\right\},u\left(1\right)=v\left(1\right)$ and $u^{\prime }\left(1\right)=v^{\prime }\left(1\right)=2$ , then find the value of $f^{\prime }\left(1\right)$ .

The focal f to a mirror is given by $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ where u and v represent object and image distance respectively then

In the mirror formula, (1)/(f)=(1)/(v)+1/u,f does not change when u is changed.

Plot the graphs of |(1)/(V)| vs |(1)/(u)| where v is image distance and u is object distance for the conditions given below:

If $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$, then make v as the subject of the formula.

The focal length of a less is given by the formula $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$. Make f as the subject of the formula.
if $u=20cm$ and v=30, then find f.
The following steps are involved in solving the above problem. Arrange them in sequential order.
(A) Given $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
(B) $\Rightarrow f=\frac{uv}{u+v}$
(C) $f=\frac{20\times 30}{20+30}=\frac{600}{50}=12$cm.
(D) $\Rightarrow \frac{1}{f}=\frac{v+u}{uv}$

उत्तल लेंस के लिए सूत्र $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ निगमित कीजिए।

कथन 1 . गोलीय दर्पण की फोकस - दूरी दर्पण से वस्तु एंव प्रतिबंब की दूरियों पर निर्भर करती हैं ।
कथन 2. गोलीय दर्पण की फोकस दूरी f, वस्तु की दूरी u एंव प्रतिबिब दूरी v के बीच निम्नलिखित संबंध होता हैं ।
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$.

Statement I : Focal length of a spherical mirror depends on the distances of object and image from the mirror.
Statement $II:\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

What is f in mirror formula (1)/(u)+(1)/(v)=(1)/(f) ?