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Given,
Size of the object, h = + 3 cm
Object distance, u = - 15 cm
Focal length, f = + 10 cm   [f is +ve for a convex lens]
By lens formula,           
                          $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} = \frac{1}{\mathrm{f}}$ 

$\therefore$                      $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{u}}+\frac{1}{\mathrm{f}} \\ = \frac{1}{-15}+\frac{1}{10} \\ = \frac{-2+3}{30} \\ = \frac{1}{30} \end{gathered}$$ 

i.e., Image distance, $\mathrm{v} = + 30 \mathrm{cm}.$

Magnification,  $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = \frac{\mathrm{v}}{\mathrm{u}}$ 

$\therefore$ Image size, $\mathrm{h}\text{'} = \frac{\mathrm{vh}}{\mathrm{u}} = \frac{(+30) (+3)}{-15} = -6 \mathrm{cm}.$

As v is +ve and h' is negative, so a real and inverted image is formed at 30 cm behind the lens. 

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Linear magnification: The linear magnification produced by a lens is defined as the ratio of the size of the image formed by the lens to the size of the object. It is denoted by m. Thus,

Convex lens:


Here,                             
                              
According to new cartesian sign convention,
                                    
                         

                      
Concave lens:
     
Here,                          
                     
According to new cartesian sign convention,
                       
                           

                                                    

                  

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We are given a convex lens.
Here,
Object size, h = + 5 cm
Focal length, f = +20 cm   [f is +ve for a convex lens]
Object distance, u = -30 cm
Image distance, v = ?
Image size = ? 


For a lens using the lens formula,

                     $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} =\frac{1}{\mathrm{f}}$

$\therefore$                 $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{u}}+\frac{1}{\mathrm{f}} \\ = \frac{1}{-30}+\frac{1}{20} \\ = \frac{-2+3}{60} \\ = \frac{1}{60} \end{gathered}$$

i.e., Image distance,  $\mathrm{v}= + 60 \mathrm{cm}$ 

Magnification,  $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = \frac{\mathrm{v}}{\mathrm{u}}$

$\therefore$$\mathrm{m} = \frac{\mathrm{v}}{\mathrm{u}} = \frac{+60 \mathrm{cm}}{-30 \mathrm{cm}} = -2.$

Also, Image size,  
                           $$\begin{gathered} \mathrm{h}\text{'} = \frac{\mathrm{vh}}{\mathrm{u}} \\ = \frac{(+60) \times (+5)}{(-30)} \\ = -10 \mathrm{cm}. \end{gathered}$$

The positive sign of v shows that the image is formed at a distance of 60 cm to the right of optical centre of the lens. Negaticve sign of h' implies that the image formed is inverted. 

Therefore, the image is real and inverted. 

Thus, a real and inverted image which is 10 cm tall, is formed at a distance of 60 cm on the right side of the lens.
The image is two times enlarged in size than the object.

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A concave lens always forms a virtual, erect image on the same side of the object. 

Given,
Image distance, v = -10 cm
Focal length,    f = - 15 cm            [f is -ve for a concave lens]
Object distance, u = ? 

Now, using lens formula, 

                        $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} =\frac{1}{\mathrm{f}}$
we have,
                       $$\begin{gathered} \frac{1}{\mathrm{u}} =\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{f}} \\ = \frac{1}{-10} -\frac{1}{-15} \\ = \frac{-3+2}{30} \\ = -\frac{1}{30} \end{gathered}$$ 

i.e.,                    $\mathrm{u} = -30 \mathrm{cm}$ 

Thus the object should  be placed at a distance of 30 cm from the lens on the left side. 

Now, 

Magnification, $\mathrm{m}= \frac{\mathrm{v}}{\mathrm{u}} = \frac{-10}{-30} = +\frac{1}{3} = + 0.33$ 

Since, magnification is positive, we can say that the image is erect and virtual.
The size of the  image is reduced to one-third in size than the object after refraction.

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Image formation by a concave lens:
(i) Object at infinity: The rays from an object at infinity are parallel to each other. The rays diverge on refraction through the lens. A virtual, erect and extremely diminished image is formed at the focus F.

Fig. Image formed by a concave lens when the object is at infinity

(ii) Object between infinity and optical centre O of the lens: A ray AN parallel to the principal axis, after refraction, appears to come from the focus F₁. Another ray from  lies between optical centre and infinity.
A passes undeviated through the optical centre O. The two rays appear to diverge from the point A'. Thus A' is the virtual image of A. Hence A'B' is the complete, virtual, erect and diminished image of the object AB. For all positions of the object the image is always virtual, erect and diminished and is formed between focus F₁ and the optical centre O.


Fig.  Image formed by a concave lens when the object